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Full text of "Elements of engineering thermodynamics"

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ELEMENTS OF 

ENGINEERING 
THERMODYNAMICS 



BY 
JAMES A. MOYER 

Director of the Massachusetts Department of University Extension 

formerly Professor of Mechanical Engineering in the 

Pennsylvania State College 

JAMES P. CALDERWOOD 

Professor of Mechanical Engineering in the Kansas State 
Agricultural College 

ANDREY A. POTTER 

Dean of Engineering at Purdue University, formerly 

Dean of Engineering at the Kansas State 

Agricultural College 



NEW YORK 

JOHN WILEY & SONS, Inc. 

London: CHAPMAN & HALL, Limited 
1920 



< 3^ 



Copyright, 1920, by 
JAMES A. MOYER, JAMES P. CALDERWOOD 

AND 

ANDREY A. POTTER 



£o-/t$te 



Manufactured in the U.S.A. 

CCT I5IS20 
©CI.A576885 



PREFACE 



This treatise is an extension of a briefer work entitled " Engi- 
neering Thermodynamics" by James A. Moyer and James P. 
Calderwood. The additions and changes are made to suit the 
needs of those who had successful experience in using the original 
publication and found it desirable to add supplementary material 
to make it sufficiently inclusive for special institutional require- 
ments. A great deal of the new material is supplied by A. A. 
Potter. 

This book is intended to bring out the fundamental principles 
of Engineering Thermodynamics, and is particularly intended 
for use in technical colleges where it is possible to give special 
courses on the subjects of steam turbines, internal combustion 
engines, refrigeration and other applications of thermody- 
namics. 

The new material includes the theory of the hot air engine 
and internal combustion engine cycles. The appendix includes 
logarithmic tables, the properties of gases, the properties of 
steam, ammonia, sulphur dioxide and carbon dioxide. 

Every engineering student should become familiar with stand- 
ard works on the subject of thermodynamics. This book should 
consequently be supplemented by references to standard works on 
this subject. A representative list of such reference books is 
given in a table of the appendix. 

The authors are particularly indebted to the following pro- 
fessors and instructors in Mechanical Engineering Departments 
who gave valuable suggestions and criticisms: Edwin A. Fessen- 
den, Pennsylvania State College; J. E. Emswiler, University of 
Michigan; G. L. Christensen, Michigan College of Mines; Roy 



in 



IV 



PREFACE 



B. Fehr, formerly of Pennsylvania State College; H. L. Seward, 
Yale University; J. R. Wharton, University of Missouri; and 
J. J. Wilmore, Alabama Polytechnic Institute. 

J. A. MOYER, 

J. P. CALDERWOOD, 

A. A. POTTER. 



CONTENTS 



Pages 
Chapter I. — Thermodynamic Principles and Definitions 1-13 

Definition, Scope and Object of Thermodynamics; Heat; Temperature; 
Thermometers; Absolute Zero; Units of Heat; Specific Heat; Pressure; 
Volume; Work; Power; Indicated Horse-power; Brake Horse-power; 
Mechanical Equivalent of Heat; First Law of Thermodynamics; Second 
Law of Thermodynamics; Effects of Heat Application; The Heat Engine; 
Thermal Efficiency of a Heat Engine; Problems. 

Chapter II. — Properties of Perfect Gases 14-25 

Perfect Gas Defined; Relation between Pressure, Volume, and Temper- 
ature of a Perfect Gas; Boyle's Law; Charles' Law; Combination of 
Boyle's and Charles' Laws; The Law of Perfect Gases; Heat and its 
Effect upon a Gas; External Work; Internal Energy; Joule's Law; 
Relation of Specific Heats and the Gas Constant; Ratio of the Two 
Specific Heats; Values of the Specific Heats; Problems. 

Chapter III. — Expansion and Compression of Gases 26-39 

Expansion at Constant Pressure; Expansion at Constant Volume; 
Expansion at Constant Temperature, or Isothermal; Adiabatic Expan- 
sion and Compression; Change of Internal Energy during Adiabatic 
Processes; Relation between Volume, Pressure and Temperature in 
Adiabatic Expansion of a Perfect Gas; Problems. 

Chapter IV. — Cycles of Heat Engines using Gas 40-61 

(1) The Carnot Cycle; Reversible Cycles. 

(2) Hot Air Engine Cycle. The Stirling Engine; The Ericsson Engine. 

(3) Internal Combustion Engine Cycles. The Lenoir Engine; The Otto 

Cycle; The Brayton Cycle; the Diesel Cycle. 
Problems. 

Chapter V. — Properties of Vapors 62-90 

Saturated and Superheated Vapors; Theory of Vaporization; Vapor 
Tables; Relation between Temperature, Pressure and Volume of Satu- 
rated Steam; Heat in the Liquid (Water); Latent Heat of Evaporation; 
External Work of Evaporation; Total Heat of Steam; Internal Energy 
of Evaporation and of Steam; Steam formed at Constant Volume; Wet 
Steam; Superheated Steam; Drying of Steam by Throttling or Wire- 



VI CONTENTS 

i • ^ • , , Pages 

drawing; Determination of the Moisture in Steam; Throttling or 

Superheating Calorimeter; Barrus Throttling Calorimeter; Separating 
Calorimeters; Condensing or Barrel Calorimeter; Equivalent Evapo- 
ration and Factor of Evaporation; Vapors as Refrigerating Media; 
Problems. 

Chapter VI. — Entropy 91-103 

Entropy Changes during Constant Pressure Expansions of Gases; 
Entropy Changes of Gases at Constant Volume; Entropy Changes 
during Isothermal Processes of a Gas; Entropy Changes during Reversible 
Adiabatic Processes of Gases; Entropy Changes during a Carnot Cycle; 
Temperature-Entropy Diagrams for Steam; Calculation of Entropy for 
Steam; Total Entropy of Steam; The Mollier Chart; Temperature- 
Entropy Diagram for the Steam Power Plant; Problems. 

Chapter VII. — Expansion and Compression of Vapors 104-116 

Expansion of Wet Steam at Constant Volume; Expansion of Super- 
heated Steam at Constant Volume; Expansion at Constant Pressure; 
Isothermal Lines of Steam; Adiabatic Lines for Steam; Quality of 
Steam during Adiabatic Expansion; Graphical Determination of Quality 
of Steam by Throttling Calorimeter and Total Heat-Entropy Diagram; 
Poly tropic Expansion (n = 1); Problems. 

Chapter VIII. — Cycles of Heat Engines using Vapors 11 7-149 

Carnot Cycle; Rankine Cycle; Practical or Actual Steam Engine Cycle; 
Efficiency of an Engine Using Steam Without Expansion; Adiabatic 
Expansion and Available Energy; Available Energy of Wet Steam; 
Available Energy of Superheated Steam; Application of Temperature- 
Entropy Diagram to Analysis of Steam Engine; Combined Indicator 
Card of Compound Engine; Hirn's Analysis; Clayton's Analysis; Prob- 
lems. 

Chapter IX. — Flow of Fluids 150-177 

Flow Through a Nozzle or Orifice; Weight per Cubic Foot; Maximum 
Discharge; Shape of Nozzle; Flow of Air Through an Orifice; Receiver 
Method of Measuring Air; Flow of Vapors; Velocity of Flow as Affected 
by Radiation; Friction Loss in a Nozzle; Impulse Nozzles; Turbine 
Losses; Reaction Nozzles; Coefficient of Flow; Injectors; Weight of 
Feed Water Supplied by an Injector per Pound of Steam; Thermal Effi- 
ciency of an Injector; Mechanical Efficiency of an Injector; Orifice 
Measurements of the Flow of Steam; Flow of Steam Through Nozzles; 
Flow of Steam when the Final Pressure is more than 0.58 of the Initial 
Pressure; Length of Nozzles; Efficiency of Nozzles; Under- and Over- 
Expansion; Non-expanding Nozzles; Materials for Nozzles; Problems. 



CONTENTS VU 

Page 
Chapter X. — Applications of Thermodynamics to Compressed Atr 

and Refrigerating Machinery 178-196 

Compressed Air; Air Compressors; Work of Compression; Effect of 
Clearance upon Volumetric Efficiency; Two Stage Compression; Re- 
frigerating Machines or Heat Pumps; Unit of Refrigeration; Systems 
of Mechanical Refrigeration; The Air System of Refrigeration; The 
Vapor Compression System of Refrigeration; The Vapor Absorption 
System of Refrigeration; Coefficient of Performance of Refrigerating 
Machines; Problems. 



SYMBOLS 



A = area in square feet, also used to represent the reciprocal of the mechanical 

equivalent of heat, y-fg-. 
B.t.u. = British thermal units (= 778 ft.-lbs.). 

Cp = specific heat at constant pressure in B.t.u. per pound per degree. 
Cv = specific heat at constant volume in B.t.u. per pound per degree. 
C = a general constant in equations of perfect gases. 

E = external work in B.t.u. per pound; also sometimes used to express effi- 
ciency, usually as a decimal. 
E a = available energy in B.t.u. per pound. 
F = force in pounds. 
H = heat per pound in B.t.u.* 
Hsup. = total heat of superheated steam, B.t.u. per pound. 
In = total internal energy of steam (above 32 F.) in B.t.u. per pound, some- 
times designated by E and i". 
I L = internal energy of evaporation of steam in B.t.u. per pound, sometimes 

designated by p. 
J = mechanical equivalent of heat = 778 (use becoming obsolete). 
K = specific heat in foot-pound units. 
L = latent heat of evaporation in B.t.u. per pound. 
P = pressure in general or pressure in pounds per square foot. 
Q = quantity of heat in B.t.u. 
R = thermodynamic constant for gases; for air it is 53.3 (in foot-pound units 

per pound.) 
T = absolute temperature, in Fahr. degrees = 460 + t. 

V = volume in cubic feet, also specific volume and velocity in feet per second. 
W = work done in foot-pounds. 
a = area in square inches. 
c = constant of integration. 
d = distance in feet. 

e = subscript to represent base of natural logarithms. 
g = acceleration due to gravity = 32.2 feet per second per second. 
h = heat of the liquid per pound in B.t.u. (above 32 ° F.) also designated 

by q and i' . 
i' = heat of liquid in B.t.u. per pound (above 32 F.). 
i" = total internal energy of steam. 
k = a constant, 
log = logarithm to base 10. 
loge = logarithm to natural base e (Naperian). 

* In steam tables it is total heat above 32° F. 
ix 



SYMBOLS 

n = general exponent for V (volume) in equations of perfect gases, also some- 
times used for entropy of the liquid in B.t.u. per degree of absolute tem- 
perature. 

p = pressure in pounds per square inch. 

q = sometimes used for heat of the liquid in B.t.u. per pound (above 3 2° F.). 

r = ratio of expansion also sometimes used for latent heat of evaporation in 
B.t.u. per pound. 

s' = entropy of the liquid. 
s" = total entropy of vapors. 

t = temperature in ordinary Fahr. degrees. 

u = difference between the specific volume of a vapor and that of the liquid 
from which it is formed. 

v = specific volume, in cubic feet per pound (in some steam tables) . 

w = weight per cubic foot = density, also used to designate weight of a sub- 
stance in pounds. 

x = quality of steam expressed as a decimal. 

r< 

y = ratio of specific heats — . 

Cp 

4> = total entropy -^f, sometimes designated by S". 

6 — entropy of the liquid in B.t.u. per pound per degree of absolute tem- 
perature, 
p = internal energy of evaporation, B.t.u. per pound. 
<r = volume of water in a cubic foot of saturated steam. 
" — inches. 



ELEMENTS OF 

Engineering Thermodynamics 



CHAPTER I 
THERMODYNAMIC PRINCIPLES AND DEFINITIONS 

Thermodynamics. Thermodynamics deals with the relation 
between heat and mechanical work. The object of the study of 
thermodynamics is to consider factors which influence the 
efficiency of heat power machinery. 

Thermodynamics makes it possible to predict the perform- 
ance of steam engines and steam turbines when operating under 
conditions of increased pressure, of higher vacuums and expan- 
sions, of higher superheats, of jacketing cylinders, of using the 
working medium in several cylinders one after the other, — that 
is, compounding instead of expanding the steam only in a single 
cylinder, — of inserting receivers between the cylinders and of 
reheating the working medium as it passes from one cylinder to 
the next. 

By means of calculations based upon the study of thermo- 
dynamics it is possible to determine the effect of increasing the 
compression pressures of a gas engine mixture before it is 
ignited, the result of incomplete cooling upon the efficiency of 
an air compressor, and similar problems. 

Another important service which the study of thermodynamics 
renders is that of showing what maximum efficiency is attain- 
able for any heat engine operating under a given set of conditions. 
It often happens that tests indicate an efficiency very much bet- 
ter than is usually obtained with any of the present types of 
engines. In such cases thermodynamic calculations will show 
conclusively whether the results secured are possible. The 
ability to interpret correctly the results of experiments performed 
on all kinds of heat engines requires a knowledge of the basic 
principles of thermodynamics. 



2 THERMODYNAMIC PRINCIPLES AND DEFINITIONS 

Heat. Heat is a form of energy and not a material substance. 
The heat of a body depends on the vibratory motion of the small 
particles or molecules of which the body is built up, the greater 
the velocity and amplitude of the vibration of these molecules, 
the higher is the temperature of the body. 

Heat can be transferred by conduction, radiation and con- 
vection. 

The transfer of heat between the different particles of the 
same body or between several bodies is called conduction. 

In a heated body the particles are in violent agitation and as 
a result waves are formed and are emitted or are radiated 
through space to other bodies. 

The transfer of heat by the motion of the heated particles is 
called convection. This phenomenon is exhibited in liquids and 
gases. Thus, when a hot body heats the air in contact with it, 
the currents of air which are produced by the process of con- 
vection ascend and are replaced by cooler air. 

Temperature. Temperature is the indication of the sensible 
heat of a substance and can be measured by a thermometer. The 
temperature does not measure the quantity of heat energy in the 
substance, but indicates only the relative heat intensity, which 
can be revealed by the senses of the observer. 

Thermometers. There are three thermometric scales: 

The Centigrade or Celsius degree is T fo of the temperature in- 
terval between the melting point of ice and the boiling point of 
water at atmospheric pressure, these two fixed points being de- 
noted o° C. and ioo° C. respectively. 

The Fahrenheit degree is T -§-o of the temperature interval 
between these two fixed points, the melting point of ice being 
taken at 3 2° F. and the boiling point of water at 212 F. 

The Reaumur scale has the melting point of ice at o° R. and the 
boiling point of water at 8o° R. 

The thermometric scales generally used are the Centigrade 
and Fahrenheit, the relations between these scales being: 

Degrees C. = | [degrees F. — 32]. (1) 

Degrees F. = f degrees C. + 32. (2) 



UNITS OF HEAT 3 

Mercury thermometers, as ordinarily constructed, have the 
space above the mercury under a vacuum. Such thermometers 
cannot be used for the measurement of temperatures exceeding 
500 F., as the vacuum reduces the boiling point of mercury. 
The range of mercury thermometers can be increased to about 
900 F. by filling the space above the mercury with some inert 
gas like nitrogen. 

For the measurement of very high temperatures thermo- 
electric pyrometers are best suited. 

Absolute Zero. In the graduation of liquid thermometers, the 
vaporization of the liquid at high temperatures and its freezing 
at low temperatures limits the thermometric range. The funda- 
mental scale of temperature measurement is based on Thom- 
son's absolute thermometric scale, which is independent of the 
nature of any thermometric substance. The zero of the abso- 
lute scale or the absolute zero is taken as a point which marks the 
absence of heat energy or of molecular vibrations of a body. 

The absolute zero is 459. 5 (practically 460 ) below the zero 
on the Fahrenheit scale and 273.0 below the zero on the Centi- 
grade scale. 

Calling the absolute temperature T and the temperature as 
measured by a thermometer /, 

On the Fahrenheit scale T = t + 460. (3) 

On the Centigrade scale T = / + 273. (4) 

Units of Heat. Heat is measured in heat units. A heat unit 
is the amount of heat required to raise the temperature of one 
unit weight of water one degree. In the English system of 
measures the heat unit is the British thermal unit (B.t.u.), which 
is defined as the amount of heat required to raise the tempera- 
ture of one pound of water one degree on the Fahrenheit scale. 
The heat unit in the metric system is the calorie * which is de- 
fined as the heat required to raise the temperature of one kilo- 
gram of water one degree on the Centigrade scale. 

* Since one kilogram = 2.204 pounds and one degree C. = |° F., 1 calorie 
= f X 2.204 = 3-968 B.t.u. 



4 THERMODYNAMIC PRINCIPLES AND DEFINITIONS 

To correctly define the British thermal unit it is necessary 
to state at what temperature the rise of one degree on the Fahren- 
heit scale is to occur, because the specific heat of water is slightly 
variable. This heat unit (B.t.u.) is sometimes denned as the 
amount of heat required to raise the temperature of water one 
degree Fahrenheit at the condition of maximum density of water, 
that is, between 39 and 40 degrees Fahrenheit. Other definitions 
are based on the amount of heat required to raise the temperature 
of water one degree Fahrenheit between 60 to 61 degrees Fahren- 
heit. Still another definition, which is generally considered to be 
the most accurate, defines a British thermal unit as one one- 
hundred-and-eightieth ( T -§~o) of the amount of heat required to 
raise the temperature of water from 32 to 212 degrees Fahrenheit. 
In other words, according to this last definition, the British ther- 
mal unit is the average value of the amount of heat required to 
raise the temperature of one pound of water one Fahrenheit 
degree between the conditions of freezing and boiling at atmos- 
pheric pressure. 

Specific Heat. As the addition of the same quantity of heat 
will not, as a rule, produce the same temperature changes in equal 
weights of different substances, it is evident that the amount of 
heat in any substance will depend on the capacity of the substance 
for heat. For this reason it is necessary to allow for the relative 
heat capacity or the specific heat (C) of a substance. Specific 
heat can be defined as the ratio of the heat added to the tempera- 
ture change produced in a unit weight of a substance. It can 
also be defined as the resistance which a substance offers to a 
change in its temperature, the resistance of water being taken as 
unity. In the English system the specific heat is the number of 
British thermal units (B.t.u.) required to raise the temperature 
of a pound of the substance one degree Fahrenheit. 

Thus if Q is the quantity of heat added to one pound of a sub- 
stance, the temperature change is, 

h-\=% (5) 

or Q = C (fe - h). 



SPECIFIC HEAT 5 

If the specific heat is a variable, 

Cdt. (6) 

The following problem illustrates the application of equation 
(6): 
The specific heat of a substance is expressed by the equation, 

C = 0.241 1 2 + 0.000009 t. 

What amount of heat is required to raise the temperature of 
one pound of the substance from o° to ioo° F.? 
Solution. Since the specific heat is variable, 






h 

Cdt. 



substituting the value of C and integrating, 

(0.241 1 2 + 0.000009 1) & 
h 

n ioo° , nn 1000 

= O.24I 1 2 [/J QO + O.OOOOO9 - 

L 2 Jo° 

= O.24I 1 2 (lOO) + O.OOOOO9 (5000) 

= 24.112 + 0.045 = 24.157 B.t.u. 

The specific heat of gases and vapors changes considerably in 
value according to the conditions under which the heat is applied. 
If the heat is applied to a gas or a vapor held in a closed vessel, 
with no change in volume, no work is performed, and, therefore, 
all the heat added is used to increase the temperature. This is 
the condition in a boiler when no steam is being drawn off. In 
this case the symbol C v represents the specific heat during heat 
application at constant volume. If, on the other hand, the heat- 
ing is done while the pressure is kept constant and the volume is 
allowed to change permitting expansion and the performance of 
work, the symbol C v is used and represents the specific heat dur- 
ing heat application at constant pressure. 



6 THERMODYNAMIC PRINCIPLES AND DEFINITIONS 

When the problem deals with w pounds of a substance instead 
of a unit weight, equation (6) becomes, 

Q = wf h Cdt. (7) 

Pressure. Force per unit of area is called pressure. Thus, 
the pressure exerted by a gas or vapor is expressed in the 
English system in pounds per square inch, pounds per square 
foot, inches of mercury or atmospheres. In the metric system, 
pressure is expressed in kilograms per square centimeter or 
millimeters of mercury. 

Gages always read pressures above atmospheric pressure or 
above vacuum. The absolute pressure is the sum of the gage 
and atmospheric pressures. Thus, if a gage reads 75 pounds 
pressure (per square inch), and the barometer is 29.65 inches of 
mercury, the atmospheric or barometric pressure is, in pounds per 
square inch, 

29.65 X 0.491 = 14.56. 

(0.491 is the weight of a cubic inch of mercury at 70 F.) 
The absolute pressure is: 

75 + 14.56 = 89.56 pounds per square inch. 

In thermodynamic equations the unit of pressure is usually 
expressed in pounds per square foot. 

Volume. By specific volume is meant the amount of space 
occupied by a unit weight of a substance, expressed in cubic feet 
or in cubic meters. Thus, the volume of one pound of steam at 
atmospheric pressure is its specific volume and is equal to 26.79 
cubic feet. 

Work. Work is done by a force during a given displacement 
and is independent of the time. The foot-pound is the unit of 
work in the English system. Thus, when a body weighing one 
pound is raised through a distance of one foot, the resulting work 
is a foot-pound. Similarly, the product of the pressure in pounds 



POWER 



per square foot and the volume in cubic feet is equal to work in 
foot-pounds. 




Space 
Fig. i. — Work Diagram. 

Work being the product of two dimensions, it may be repre- 
sented graphically by the area of a closed figure, the coordinates 
of which are force and distance, or pressure and volume. Thus, 
A (Fig. i ) represents the position of a body between F (force) and 
5 (space) coordinates, being the origin or starting point. The 
distance A D f rpm the OS line represents the force against which 
it is acting and AA', measured from the OF line, is the distance it 
has moved from the starting point. If the body moves from A 
to B along the path AB, the area A BCD represents the work 
done. 

If the equation of the path is known, the work is 



W 



-£" 



SB 

FdS. 



(8) 



Power. Power is the rate of doing work, or the work done 
divided by the time required to do it. In the English system, the 
unit of power is the horse power. It is the power required to 
raise 550 pounds through a vertical distance of one foot in one 



8 THERMODYNAMIC PRINCIPLES AND DEFINITIONS 

second, or 33,000 pounds one foot in one minute. To obtain the 
horse power, the work in foot-pounds per minute must be obtained 
and the result divided by 33,000. In the metric system, the unit 
of power is the watt. One horse power is equal to 746 watts. 
The French horse power (cheval) is 542.5 foot-pounds per second. 

Indicated Horse Power. The term indicated horse power is 
applied to the rate of doing work by a gas or a vapor in the cylin- 
der of an engine. It is obtained by means of an engine indicator. 

If P represents the mean effective pressure (average unbal- 
anced pressure) in pounds per square inch, as shown by the 
indicator card, and A the effective area of the piston in square 
inches, the total pressure exerted on the piston is PA. If the 
piston has a stroke of L feet, the work per stroke is PAL and the 
work per minute is PA L N, where N represents the number of 
revolutions per minute. The indicated horse power is 

T , Work per minute PA LN 

I.h.p. = £ = -. (9) 

33,000 33>°°° 

Brake Horse Power represents the actual power which an 
engine can deliver for the purposes of work. The difference 
between indicated and brake horse power of an engine represents 
the horse power lost in friction. The brake horse power can be 
measured by some form of friction brake, which absorbs the power 
measured, and is called an absorption dynamometer, or by a 
transmission dynamometer. In either type of dynamometer, 
if F is the effective pull in pounds, L the lever arm, in feet, 
through which the weight is exerted, and N the number of revo- 
lutions of the shaft per minute, the brake horse power is 

-,, 2TFLN , , 

B.h.p. = (10) 

33>°°° • 

Mechanical Equivalent of Heat. There is a definite quantita- 
tive relation between work expended and heat produced. This 
relation between heat and work is called the mechanical equiva- 
lent of heat and is designated by /. In the English system, 

J = 778 foot-pounds 
or 1 B.t.u. =778 foot-pounds. 



EFFECTS OF HEAT APPLICATION 9 

In the metric system, 

/ = 427 kilogrammeters 
or 1 calorie = 427 kilogrammeters. 

The reciprocal of J, or the heat equivalent of work, is designated 

by A. where A = — - in the English system or in the metric 

778 6 427 

system. 

First Law of Thermodynamics. The statement of the definite 
relation between heat and mechanical work constitutes what is 
known as the first law of thermodynamics. It is usually 
expressed: 

"Heat and mechanical energy are mutually convertible and 
heat requires for its production and produces by its disappear- 
ance mechanical work in the ratio of the mechanical equivalent 
of heat." In other words, this law is a statement of the conserva- 
tion of energy as regards the equivalence of mechanical work and 
heat. 

Second Law of Thermodynamics. "In order to transform the 
heat of a body into work, heat must pass from that body into 
another at a lower temperature." Thus, there must be a dif- 
ference of level in the transformation of heat energy into work 
and heat cannot be transformed from one body to another at a 
higher temperature, unless work is expended in order to produce 
such a transfer of heat. 

This law states as regards heat engines the limits to their pos- 
sible performance, which would be otherwise unlimited, if only 
the "first law" of thermodynamics is considered. It means, also, 
that no heat engine converts or can convert into work all of the 
heat supplied to it. A very large part of the heat supplied is 
necessarily rejected by the engine in the form of unused heat. 

Effects of Heat Application. If a quantity of heat Q is im- 
parted to a body, the following effects will be produced: 

1. The temperature of the body will rise; 

2. The volume of the body will increase; 

3. The body will be capable of doing external work. 



IO THERMODYNAMIC PRINCIPLES AND DEFINITIONS 

Representing the above effects by S, V and W, 

Q = S+V + W, (n) 

in which S represents that quantity of heat, termed the sensible 
heat, which was utilized in raising the temperature of the body; V 
represents that quantity of heat which was absorbed by the body 
in increasing its store of internal energy other than that associated 
with the sensible heat; and W is the heat which was absorbed to 
perform the external work in increasing the volume of the body 
against the resistance offered by external substances. S + V 
represents that portion of the heat that was chargeable to the 
increase of internal work. This may be termed the intrinsic 
energy increase and designated by /. 
Thus, for the addition of an infinitesimal quantity of heat, 

dQ = dI + dW 






or Q = | ~dl - f *PdV. (12) 



In equation (12), which represents the effect of heat application, 
the intrinsic energy change / dl depends on the physical state 

of the body, that is, whether it is a solid, a liquid, a vapor, or a 

/■» V-t 
P dV depends on the char- 
Vi 

acter of the path in the work diagram. 

The Heat Engine. The heat engine is a machine which con- 
verts the heat energy of solid, liquid or gaseous fuel into work. 
This conversion depends on the variation in the pressure, volume 
and temperature of a gas or a vapor and can be accomplished in 
two ways: 

First, by "external combustion," in which case the fuel is 
burned outside of the engine cylinder; the heat developed by the 
combustion of the fuel is conducted to the working substance or 
heat medium through walls; this working substance does work 
on a piston in the case of a reciprocating engine or on a vane or 



THERMAL EFFICIENCY OF A HEAT ENGINE n 

" blade " in a turbine. To this class belong steam engines of the 
reciprocating, turbine or rotary types and also external-combus- 
tion hot air engines. Thus, in the case of the steam engine, the 
fuel, which may be coal, wood, petroleum or gas, is burned out- 
side of the boiler shell and the resulting heat is transmitted by 
conduction through the metal of the shell to the working sub- 
stance, which is water. When enough heat has been added to the 
water to produce a change in its physical state, water vapor or 
steam is formed at the required pressure. This vapor, which 
may be dry, wet or superheated, if allowed to act on the piston 
of the engine, will do work. 

Another method of converting heat into mechanical energy is 
by burning the fuel rapidly or slowly inside of an engine cylinder 
or in a communicating vessel, the products of combustion being 
allowed to act directly on the piston of the engine. To this class 
belong gas, petroleum and alcohol engines which are called 
" internal combustion " engines. 

Thermal Efficiency of a Heat Engine. By thermal efficiency 
(E) is meant the ratio of the heat converted into work (AW) to 
the heat supplied by the engine (Qi), or 

77 AW 

E - -^-. (13) 

Since only a part of the heat supplied to an engine can be con- 
verted into work, the above ratio is a fraction always less than 
unity. 

PROBLEMS 

1. Give examples of the transfer of heat by conduction, radiation and 
convection. 

2. Show that the kilogram calories per kilogram X 1.8 give B.t.u. per 
pound. 

3. Prove that the product of a pressure in pounds per square foot and a 
volume in cubic feet results in foot-pounds of work. 

4. If the specific heat of a substance is 0.65, how many B.t.u. are re- 
quired to raise the temperature of 10 lbs. of the substance through io° F.? 

5. The specific heat of a substance is expressed by the equation, 

C = O.S — O.C2 t. 



12 THERMODYNAMIC PRINCIPLES AND DEFINITIONS 

What heat is required to raise the temperature of 5 lbs. of the substance 
from io° to ioo° F.? 

6. Convert — 40 C. into degrees Fahrenheit. 

7. Change 350 F., 212 F. and 160 C. to absolute Fahrenheit tempera- 
tures. 

8. Prove that the weight of 1 cu. in. of mercury is equal to 0.491 lb. 

9. If the barometric reading is 29.2 ins., change 140 lbs. per sq. in. gage 
and also 27 ins. vacuum into pounds per square inch absolute pressure. 

10. Calculate the indicated horse power of a 12" X 13" steam engine 
which operates at a speed of 265 r.p.m. The mean effective pressure of the 
head end is 27.5 lbs. per sq. in. and of the crank end 27.8 lbs. per sq. in. The 
diameter of the piston rod is 1 f inches. 

11. Calculate the brake horse power developed by an engine as measured 
by a Prony brake, the effective pull being 32 lbs. at 250 r.p.m. and the 
lever arm 32 inches long. 

12. One pound of fuel has a heating value of 14,500 B.t.u. How many 
foot-pounds of work is it capable of producing, if all this heat is converted 
into work? 

13. An engine developed 15,560 foot-pounds of work. How much heat 
in B.t.u. was theoretically required? 

14. A heat engine receives 100,000 B.t.u. of heat in the form of fuel and 
during the same period 30,000 B.t.u. are converted into work. What per- 
centage (thermal efficiency) of the heat received by the engine was converted 
into work? 

15. A gas engine receives 20,000 B.t.u. of heat in the form of fuel and 
during the same period 3,11 2,000 foot-pounds of work are developed. What 
is the thermal efficiency of the engine? 

16. It is claimed that a certain motor generates 300,000 foot-pounds of 
work per hour and during this period receives 400 B.t.u. of heat in the form 
of fuel. Are such results possible? 

17. An oil engine uses 0.74 lb. of fuel per b.h.p. per hour. Calculate the 
thermal efficiency of this engine if the oil has a calorific value of 18,600 B.t.u. 
per lb. 

18. An engine receives 200 B.t.u. of heat per minute and exhausts during 
the same period 100 B.t.u. If no losses of heat occur within the cylinder, 

(a) How many B.t.u. of heat are being transformed into work? 

(b) What number of foot-pounds does this heat produce? 

(c) What horse power is being developed? 

(d) What is the thermal efficiency of the engine? 

19. In the manufacture of certain explosives, acids are mixed with an 
oxidizable substance. During the process the mixture must be constantly 
agitated by a stirring mechanism to maintain uniform conditions, and the 



PROBLEMS 13 

temperature of the mixture must be kept below a certain predetermined 
value to prevent explosion. If during the process of manufacture 5000 
foot-pounds of work are delivered per minute to the agitator, and 1000 B.t.u. 
are generated during the same period by the chemical reaction, how much 
heat must be absorbed per hour to maintain the temperature of the mixture 
constant? 

20. Prove that one horse power developed for one hour is equivalent to 
the consumption of 2545 B.t.u. of heat in the same period. 



CHAPTER II 



PROPERTIES OF PERFECT GASES 



In the study of thermodynamics, the working substance, or 
heat medium, through which the heat engine converts heat into 
work, is in the condition of a perfect gas or vapor. The laws 
governing the action of the two classes of substances differ. For 
this reason the study of thermodynamics is divided into the 
thermodynamics of gases and of vapors. 

When the term " gas " is used it refers to what is more properly 
called a perfect gas. A perfect gas may be defined as a fluid 
which remains in the gaseous state when subjected to changes in 
pressure or in temperature. Gases which are near their point 
of condensation are not perfect gases. Oxygen, hydrogen, nitro- 
gen, air and carbon dioxide are practical examples of perfect 
gases. 

Vapors are fluids which are readily transformed into liquids by 
a very moderate reduction in temperature or increase in pressure. 

Common examples of vapors are 
steam and ammonia. 

Relation between Pressure, Vol- 
ume and Temperature of a Per- 
fect Gas. In practically all heat 
engines, work is done by changes 
of volume of a fluid, and the 
amount of work performed depends 
only on the relation of pressure to 
volume during such change and 
not at all on the form of the vessel 
containing this fluid. 
Figure 2 shows a vessel containing a perfect gas and surrounded 
by a jacket filled with cracked ice. Its temperature will, there- 

14 



Connection to Air Pump 



WJUIMEWa 




Fig 



— Constant Temperature 
Apparatus for Demonstrating 
Relation between Pressure and 
Volume of a Gas. 



BOYLE'S LAW 1 5 

fore, be at 32 F. This vessel has a tightly fitting piston P of 
which the lower flat side has an area of one square foot. In the 
position shown the piston is two feet from the bottom of the 
vessel, so that the volume between the piston and the bottom of 
the vessel is two cubic feet. The pressure on the gas is that due 
to the piston and the weights shown. Assume this total weight 
is 100 pounds and that the air pump connected to the top of the 
vessel maintains a vacuum above the piston. Then the pressure 
on the gas below the piston is 100 pounds per square foot. If now 
the weights are increased to make the pressure on the gas 200 
pounds per square foot the piston will sink down until it is only 
one foot from the bottom of the vessel, provided the ice keeps a 
constant temperature. If the temperature is not maintained 
constant, because of the tendency of gases to expand with in- 
crease in temperature, it will be necessary to apply a total weight 
greater than 200 pounds to reduce the volume to one cubic foot. 
Similarly, if the weight on the gas were reduced to 50 pounds and 
the vessel were made high enough, the lower side of the piston 
would then be four feet from the bottom of the vessel. 

Examination of the above figures shows that if the temperature 
is constant the product of pressure and volume is a constant, and 
in this particular case it is always equal to 200 foot-pounds. 
These facts are expressed by Boyle's Law which can be stated as 
follows : 

Boyle's Law. If a unit weight of gas is compressed or ex- 
panded at constant temperature, the pressure varies inversely as 
the volume, or the product of pressure and volume remains a 
constant. Thus, if P h V± are the initial pressure and volume and 
P 2 , V 2 the final pressure and volume, 

PiV 1 = P 2 V 2 . (14) 

The laws of thermodynamics dealing with volume and pressure 
changes corresponding to temperature variations may be stated 
as follows: 

(1) Under constant pressure the volume of a given mass of 
gas varies directly as the absolute temperature. 



16 PROPERTIES OF PERFECT GASES 

(2) Under constant volume the absolute pressure of a given 
mass of gas varies directly as the absolute temperature. 

These fundamental principles, often called Gay-Lussac's or 
Charles* Laws, may also be stated thus: 

V T 
With pressure constant, — - = — -, (15) 

V 2 T 2 
p j» 
With volume constant, — - = — -, (16) 

"2 1 2 

where V\ and V 2 are respectively the initial and final volumes, 
Pi and P 2 are the initial and final absolute pressures, and 7\ and 
T 2 are the absolute temperatures corresponding to the pressures 
and volumes of the same subscripts. 

The following problem shows applications of Charles' laws: 
A gas has a volume of 2 cubic feet, a pressure of 14.7 pounds per 
square inch absolute and a temperature of 6o° F. 

(a) What will be the volume of this gas if the temperature is 
increased to 120 F., the pressure remaining constant? 

(b) What will be the pressure if the temperature is increased 
as in (a) but the volume remains constant? 

Solution, (a) Since the pressure remains constant and the 
substance is a gas, the volume varies directly as the absolute 
temperature. 

Letting Vi and T\ be the initial conditions and V 2 and T 2 be 
the final conditions, then 

60 + 460 
120 + 460' 
V 2 = 2.23 cubic feet. 

(b) Since the volume remains constant, 

Pi T\ 14.7 60 + 460 

or 



V, 


T, 




2 






or 





v 2 


T 2 




v 2 



P 2 T 2 P 2 120 + 460* 

P 2 = 16.39 pounds per square inch absolute. 

Combination of Boyle's and Charles' Laws. Equations (14), 
(15), and (16) cannot often be used as they stand, because it 



COMBINATION OF BOYLE'S AND CHARLES' LAWS 17 

does not often happen that any one of the three variables (P, V 
and P) remains constant. A more general law must be developed, 
therefore, allowing for variations in all of the terms P, V and P. 
This is accomplished by combining the above equations. 

Assume a pound of gas of which the initial conditions of pres- 
sure, volume and temperature are represented by P h Vi and T 1} 
while the corresponding final conditions are given by P 2 , V 2 and 
P 2 . The first step is in changing the volume from Vi to V 2 and 
the pressure from Pi to some intervening pressure Pi while the 
temperature Pi remains constant. This change can be expressed 
by Boyle's law (equation 14). 

With constant temperature (Pi), 



from which, by solving, 



V, Pi 

v 2 = pT (i7) 



Pi = ^ ( l8 ) 



where Pi is the resulting pressure of the gas when its volume is 
changed from Vi to V 2 , with the temperature remaining constant 
at Pi. 

The second step is in the change in pressure from Pi to P 2 and 
in temperature from Pi to T 2 , while the volume remains con- 
stant at V 2 . This step is expressed as follows: 

With constant volume (T 2 ), 

Pi Pi / \ 

T, = T,' (l9) 



which may be written 



D P 2 T 2 , x 

P2 = ~y— (20) 



Substituting now the value of Pi from (18) in (20), we have 

P > = -F5T.' (2l) 



1 8 PROPERTIES OF PERFECT GASES 

which may be arranged to read, 

P1V1 P2V2 



(22) 



The following problem shows the application of equation (22): 
A quantity of air at atmospheric pressure has a volume of 
2000 cubic feet when the barometer reads 28.80 inches of mercury 
and the temperature is 40 C. What will be the volume of this 
air at a temperature of o°C. when the barometer reads 29.96 
inches of mercury? 

Solution. Volume, pressure and temperature vary in this case 
as in the following equation, 

P1F1 P 2 V 2 



then 



T x T 2 

Letting P h Vi, T\ = initial conditions, 

P2, V 2) T 2 = final conditions, 
28.80 X 2000 29.96 X V 2 



40 + 273 o + 273 

V 2 = 1676 cubic feet. 



Now, since P 2 , V 2 and T 2 in equation (22) represent any 
simultaneous condition of the gas, we may also write the follow- 
ing more general relations: 



PiVx P 2 V 2 P 3 V 



= a constant, (23) 



2\ T 2 T 3 

and, therefore, 

PV = RT, (24) 

PV 

where R is the " gas constant," and is equal to — - 

The Law of Perfect Gases. Equation (24) expresses the law 
connecting the relation between pressure, volume and tempera- 
ture of a perfect gas. In this equation V is the specific volume, 
or the volume occupied by a unit weight of a gas at the absolute 
pressure P and absolute temperature T; R is the gas constant 



THE LAW OF PERFECT GASES 19 

in foot-pounds and depends on the density of the gas and on the 
units of measurement adopted. 

By means of equation (24) if the pressure, volume and tem- 
perature of a gas for one given condition are known, the value of 
R can be determined. 

Example 1. If the volume of air at freezing point and atmos- 
pheric pressure is 12.39 cubic feet per pound, calculate the value 
of R in the English units. 

Solution.. R = M* = I4 -7 X 144 X 12.39 _ 5 . 

To 32 + 460 

Example 2. If one pound of air occupies 5 cubic feet at a 
temperature of 200 F., find the corresponding pressure. 
Solution. Using the value of R for air as calculated above 

„ RTi ^.3 (460 + 200) , 

Px = — — = °° ° v ^ - = 7040 pounds 

Vi 5 

per square foot absolute, which is the same as 34.2 pounds per 
square inch gage pressure. 

In calculating R by equation (24) care must be taken not to 
confuse the units of measurement. It must also be remembered 
that the method, as illustrated in the above examples, gives the 
value of R for one pound of air, or for one unit weight of the gas 
in question; for w pounds, the value of the constant is wR, or: 

PV = wRT, (25) 

where P = absolute pressure in pounds per square foot, 
V = volume in cubic feet, 
w = weight of gas in pounds, 

R = the "gas constant" for one pound of gas in foot- 
pound units, 
T = the absolute temperature in Fahrenheit degrees. 

This equation is applicable to any perfect gas within the limits 
of pressure and temperature employed in common engineering 
practice. The " thermodynamic" state of a gas is known when 
its pressure, volume, temperature, weight and composition are 
known; when any four of these quantities are known the fifth can 
be found by equation (25). 



20 PROPERTIES OF PERFECT GASES 

Heat and Its Effect Upon a Gas. In equation (12) it was 
shown that in general the effect of adding heat upon a substance 
was to increase the intrinsic energy and to overcome the external 
resistances producing work. This law can be stated as follows: 

Heat supplied = increase in intrinsic energy + the external 
work done. 

The converse of this statement is equally true. The heat 
abstracted from a gas equals the decrease in intrinsic energy plus 
the negative work done. 

External Work. The external work or the work done by a 
gas in its expansion is represented graphically by Fig. 3. The 

area under the expansion line BC 



^/w//w//m;w;mm//m/z 



Pi 


i 
1 
I 


• 


1 
1 
1 


B 


C 


Sm 








3 








!/J 








Cfl 








V 








u 
Qui 









is proportional to the work done 
"^ in the expansion. If the initial 
condition of the gas at B as re- 
gards pressure and volume is rep- 
resented by Pi and Vi and the 
final condition at C by Pi and V 2 
(expansion being at constant pres- 
Vi V 2 sure), the force P moved through 

Volume / N 

a distance of abscissas (k 2 — m) 
Fig. 3. -External Work of Expansion. is & measure of the WQrk done> 

Obviously the area under the line BC divided by the horizontal 
length (V 2 — Vi) is the average value of the force P. If, further, 
and in general, we represent the area under BC by the symbol A, 
then we can write, 

— — = average value of P, 

V 2 — ML 

whether or not P is constant. And also, 

Work done = A T _ X (7 2 - Fi) = A. 

The same principle applies whether the line BC is a straight line, 
as shown in Fig. 3, or a very irregular curve, as will be shown 
later. 



INTERNAL ENERGY 21 

Internal Energy. The heat energy possessed by a gas or 
vapor, or the heat energy which is contained in a gas or vapor 
in a form similar to " potential " energy, is called its internal 
energy. It is also called intrinsic energy, since it may be said to 
" reside " within the substance and has not been transferred 
to any other substance. Thus, an amount of heat added to a 
substance when no work is performed is all added to the internal 
energy of that substance. On the other hand, when heat is added 
while work is being performed, the internal energy is increased 
only by the difference between the heat added and the work 
done. 

Internal energy may also be denned as the energy which a gas 
or vapor possesses by virtue of its temperature, and for one 
pound of gas may be expressed as follows : 

Internal energy '= C v dT (in B.t.u.), 

where T is the absolute temperature and C v the specific heat at 
constant volume. From the paragraph on specific heat it will be 
remembered that C v takes into account only that heat required 
to raise the temperature, since under constant volume conditions 
no external work is done; and, therefore, in dealing with internal 
energy, C v is always used. 

Increase in internal energy in B.t.u. (for one pound of a gas) 

= C v (r 2 - 2\). (26) 

Joule's Law. In the case of ideally perfect gases, such as 
thermodynamic equations must deal with, it is assumed, when 
a gas expands without doing external work and without taking 
in or giving out heat (and, therefore, without changing its stock 
of internal energy), that its temperature does not change. It was 
for a long time supposed that when a gas expanded without doing 
work, and without taking in or giving out heat, that its tempera- 
ture did not change. This fact was based on the famous experi- 
ments of Joule. Later investigations by Lord Kelvin and Linde 
have shown that this statement is not exactly correct as all known 
gases show a change in temperature under these conditions. This 
change in temperature is known as the " Joule-Thomson" effect. 



22 PROPERTIES OF PERFECT GASES 

Relation of Specific Heats and the Gas Constant. If heat is 
added at constant pressure, then 

Q = wC P (T 2 - 7\). (27) 

Also, by equation (26), the increase in internal energy when 
heat is added 

= wC,(T 2 - 7\). (28) 

External work = P (V 2 — Vi) foot-pounds 

- P (F2 7 Fl) B.t.u. (2 9 ) 

778 yy 

Since the heat added = increase in internal energy + external 
work, 

wc p (t 2 - ro = wc v (t 2 - r x ) + ISXi^JA. ( 3o) 

778 

By equation (25), 

P 2 V 2 = wRT 2 and P X V X = wRTl 
Substituting these values in (30), 



wC p (T 2 - 


- Ti) = wC v (T 2 — Ti) + w — 


778 


- (31) 


Simplifying, 


C = C 4- • 




(32) 




C P -C =—= AR. 

778 




(33) 



Equation (33) shows that the difference between the two 
specific heats is equal to the gas constant R, which when measured 
in foot-pounds represents the external work done by one pound 
of a gas when its temperature is increased by one degree under 
constant pressure. 

Example. The specific heat of air at constant pressure (C p ) 
is 0.2375 B.t.u. and R = 53.3 foot-pounds. Calculate the spe- 
cific heat at constant volume (C») . 

Solution. 

C v = C p — AR = 0.2375 — ^|- = 0.1690 B.t.u. 

778 



RATIO OF THE TWO SPECIFIC HEATS 



2 3 



Ratio of the Two Specific Heats (^r). The constant repre- 
senting the ratio of the two specific heats of a perfect gas is 
represented by 7 where 

n/ = — =- J ( \ 

7 C v r AR AR K34) 

Example. Calculate the value of 7 for air. C v = 0.2375 B.t.u. 
R = 53.3 foot-pounds. 



Solution. 7 = r— = = 1.40s. 

j -AR T 53-3 

C p 778 x 0.2375 

The constant designating the relation between the specific 
heats of air is often designated by k instead of 7. 

Values of the Specific Heats. Regnault after carrying on 
experiments on the specific heat of hydrogen, oxygen, air and 
•carbon dioxide concluded that all gases have a constant specific 
heat under varying pressures and temperatures. Recent experi- 
ments ' tend to show that the specific heats of substances vary 
with the pressure and temperature. The variability in the values 
of the specific heats does not influence to any very great extent 
most thermodynamic computations. 

In the application of thermodynamics to internal combustion 
engines the exact values of the specific heats are of considerable 
importance. 

Tables 1 and 2 in the appendix give constants for various gases. 



PROBLEMS 

1. Air at constant pressure with an initial volume of 2 cu. ft. and temper- 
ature of 6o° F. is heated until the volume is doubled. What is the resulting 
temperature in degrees Fahrenheit? 

2. Air is cooled at constant volume. The initial pressure is 30 lbs. per 



24 PROPERTIES OF PERFECT GASES 

sq. in. absolute and the initial temperature is ioi° F. The final condition 
has a temperature of 50 F. What is the final pressure? 

3. One pound of hydrogen is cooled at constant pressure from a volume 
of 1 cu. ft. and temperature of 300 F. to a temperature of 6o° F. What is 
the resulting volume? 

4. A tank whose volume is 50 cu. ft. contains air at 105 lbs. per sq. in. 
absolute pressure and temperature of 8o° F. How many pounds of air does 
the tank contain? 

5. An automobile tire has a mean diameter of 34 inches and a width of 
4 inches. It is pumped to 80 lbs. per sq. in. gage pressure at a temperature 
of 6o° F.; atmospheric pressure 14.6 lbs. per sq. in. absolute. 

(a) How many pounds of air does the tire contain? 

(b) Assuming no change of volume, what would be the gage pressure 
of the tire if placed in the sun at ioo° F.? 

6. A gas tank is to be made to hold 0.25 lb. of acetylene when the pres- 
sure is 250 lbs. per sq. in. gage, atmospheric pressure 14.4 lbs. per sq. in. 
absolute, and the temperature of the gas 70 F. What will be its volume in 
cubic feet? 

7. A quantity of air at a temperature of 70 F. and a pressure of 15 lbs. 
per sq. in. absolute has a volume of 5 cu. ft. What is the volume of the 
same air when the pressure is changed at constant temperature to 60 lbs. 
per sq. in. absolute? 

8. How many pounds of air are required for the conditions in problem 7? 

9. The volume of a quantity of air is 10 cu. ft. at a temperature of 6o° F. 
when the pressure is 15 lbs. per sq. in. absolute. What is the pressure of 
this air when the volume becomes 60 cu. ft. and the temperature 6o° F.? 

10. How many pounds of air are required for the conditions in problem 9? 

11. A tank contains 200 cu. ft. of air at a temperature of 6o° F. and under 
a pressure of 200 lbs. per sq. in. absolute. 

(a) What weight of air is present? 

(b) How many cubic feet will this air occupy at 14.7 lbs. per sq. in. 
absolute and at a temperature of ioo° F.? 

12. The volume of a quantity of air at 70 F. and at a pressure of 14.2 lbs. 
per sq. in. absolute is 20 cu. ft. What is the temperature of this air when 
the volume becomes 5 cu. ft. and the pressure 80 lbs. per sq. in. absolute? 

13. If the specific heat of carbon dioxide under constant pressure C v is 
0.2012 and the value of R is 35.10, find the value of the specific heat under 
constant volume C v . 

14. How many B.t.u. are required to double the volume of one pound of 
air at constant pressure from 50 F. 

15. A tank filled with 200 cu. ft. of air at 15 lbs. per sq. in. absolute and 
6o° F. is heated to 150 F. 



PROBLEMS 25 

(a) What will be the resulting air pressure in the tank? 

(b) How many B.t.u. will be required to heat the air? 

16. A tank contains 200 cu. ft. of air at 6o° F. and 40 lbs. per sq. in. abso- 
lute. If 500 B.t.u. of heat are added to it, what will be the resulting pres- 
sure and temperature? 



CHAPTER III 
EXPANSION AND COMPRESSION OF GASES 

The equation of the perfect gas in the form PV = wRT for the 
expansion or compression of gases has three related variables, 
(i) pressure, (2) volume and (3) temperature. For a given 
weight of gas with any two of these variables known the third is 
fixed. As regards the analysis of the action of heat engines, the 
pressure and volume relations are most important, and graphical 
diagrams, called pressure-volume or P-V diagrams, are frequently 
needed to assist in the analysis. The indicator diagram is a pres- 
sure-volume diagram drawn autographically by the mechanism 
of an engine indicator. A pressure- volume diagram is shown 
in Fig. 4, in which the vertical scale of coordinates represents 

pressures and the horizontal, 
^ volumes. Assume that the pres- 

'^'m 1 ^ V3 7 Vl ~ v * sure and volume of a pound of a 

« gas are given by the coordinates 

P and Vi, which are plotted in 

the middle of the diagram. It 

& 24 6 8 io will be assumed further that 

Volume, eu. ft. 

the pressure remains constant in 

Fig. 4. — Diagram of Expansion and , , -, i • 1 • 1 

~ . . r, , . t> the changes to be indicated. 

Compression at Constant Pressure. to 

Now if the gas is expanded until 
its volume becomes V 2 , then its condition as regards pressure 
and volume would be represented by P Vi. If, on the other hand, 
the gas had been compressed while a constant pressure was 
maintained, its final condition would be represented by the point 
PVz to the left of PV\. Similarly, any line whether straight 
or curved extending from the initial condition of the gas at PV\ 
will represent an expansion when drawn in the direction away 

26 



CD 



EXPANSION AT CONSTANT PRESSURE 27 

from the zero of volumes and will represent a compression when 
tending toward the same zero. 

It has been shown (page 20) that areas on such diagrams repre- 
sent the product of pressure and volume, and, therefore, work or 
energy. Thus in Fig. 4 the area under the curve PV 3 to PV 2 
represents on the scales given 100 (pounds per square foot) X 
(9 — 1) cubic feet or 800 foot-pounds irrespective of whether it is 
an expansion or a compression from the initial condition. 

Most of the lines to be studied in heat engine diagrams are 
either straight or else they can be exactly or approximately 
represented by an equation in the form 

PV n = a, constant, (35) 

where the index n, as experimentally determined, has varying 
numerical values, but is constant for any one curve. When 
the lines of the diagram are straight the areas of simple rectangles 
and triangles need only be calculated to find the work done. 
The two most common forms of curves to be dealt with in ex- 
pansions are (1) when there is expansion with addition of heat 
at such a rate as to maintain the temperature of the gas con- 
stant throughout the expansion. Such an expansion is called 
isothermal. The other important kind of expansion (2) occurs 
when work is done by the gas without the addition or abstraction 
of heat. To do this work some of the internal heat energy 
contained in the gas must be transformed in proportion to the 
amount of work done. Such an expansion is called adiabatic. 

The following problems show the application of the foregoing 
principles to various types of expansions: 

1. Expansion at Constant Pressure. One pound of air having 
an initial temperature of 6o° F. is expanded to ioo° F. under 
constant pressure. Find 

(a) External work during expansion; 

(b) Heat required to produce the expansion. 

Solution. The heat added equals the increase in internal 
energy plus the external work done. In solving for the heat 
added or required during any expansion it is only necessary to 



28 EXPANSION AND COMPRESSION OF GASES 

find the external work (which is equal to the area under the ex- 
pansion curve) and add to it the heat needed to increase the 
internal energy. 

The external work = W = P l (V 2 -V 1 ). (36) 

Its equivalent is: wR (T 2 — 7\). (37) 

W = 1 X 53.3 [(100 -f- 460) — (60 + 460)] = 2i32ft.-lbs. 

The increase in internal energy: 

I = wC v (T 2 - TO (38) 

/ = I X 0.169 [(100 + 460) — (60 + 460)] 
= 6.76 B.t.u. 

Heat required (Q) = 6.75 H — = 9.50 B.t.u. 

778 

Another method of computing the heat required to produce 
the expansion is: 

e = ^c P (r 2 -r 1 ). (39) 

Q = 1 x 0.237 [( IO ° +.460) — (60 + 460)] 
= 9.50 B.t.u. approximately. 

2. Expansion at Constant Volume. One pound of air having 
an initial temperature of 6o° F. is heated at constant volume until 
the final temperature is 120 F. Find: 

(a) External work; 

(b) Heat required. 

» 
Solution. 

Heat added (Q) = increase in internal energy -f- external work. 

External work = o. 

Then 

Heat added = increase in internal energy + o 
= wC v (T 2 - Ti) + o 

= 1 X 0.169 [(100 + 460) — (60 + 460)] + o 
= 6.76 B.t.u. 

3. Expansion and Compression at Constant Temperature 
(Isothermal). In an isothermal expansion or compression the 



ISOTHERMAL EXPANSION AND COMPRESSION 



29 



temperature of the working substance is kept constant through- 
out the process. The form of the isothermal curve on pressure- 
volume coordinates depends upon the substance. In the case 
of perfect gases Boyle's Law (equation 14) applies and we have: 



PV = C = a constant. 



(40) 



Equation (40) is that of a rectangular hyperbola. It is the 
special case of the general equation PV n = constant (35), in 
which the index n = 1 and is represented by Fig. 5. 




Volume dV 

Fig. 5. — Work done during Isothermal Expansion and Compression. 



The external work performed is shown graphically by the 
shaded area under the curve between A and B (Fig. 5). The 
two vertical lines close together in the figure are the limits of a 
narrow closely shaded area and indicate an infinitesimal volume 
change dV. Work done during this small change of volume is, 
the 

JW = P dV, 

and for a finite change of volume of any size as from V\ to V 2 
the work done, W (foot-pounds), is: 



W 






PdV. 



(41) 






30 EXPANSION AND COMPRESSION OF GASES 

For the integration of this form it is necessary to substitute P 
in terms of V. Assume that P and V are values of pressure 
and volume for any point on the curve of expansion of a gas of 
which the equation is 

PV = C. 

Then P = £• 

Substituting this value of P in equation (41), 

tv *dV 

Vl v 

W = C (log. V 2 - loge FO. (42) 

Since the initial conditions of the gas are Pi and Vi, we have 

PV = C = PxFi, 
and substituting this value of C in equation (42), we obtain 
W = PxFi (log, F 2 - log e 70 

or T7 = P1V1 loge :==? (in foot-pounds). (43) 

r 1 

Units of weight do not enter in equations (42) and (43). For 
a certain weight of gas under the same conditions, since 

Vo P 

P1V1 = wRT (in foot-pounds) and — = -=^> 

Vi P% 
then the work for w pounds is : 

V P 

W = wRT loge — = wRT log e -5^ (in foot-pounds). (44) 

V I P2 

. V* 

Often the ratio — is called the ratio of expansion and is rep- 

V\ 
resented by r. Making this substitution we have, 

W = wRT log s r. (45) 

Equations (42), (43), (44) and (45) refer to an expansion from 
P1V1 to P2V2. If, on the other hand, the work done is the result 
of a compression from P1V1 to P3V3 the curve of compression 



ISOTHERMAL EXPANSION AND COMPRESSION 31 

would be from A to C and the area under it would be its graphical 
representation. Equations (43), (44) and (45) would represent 
the work done for compression the same as for expansion, except 
that the expression would have a negative value; that is, work is 
to be done upon the gas to decrease its volume. 

The isothermal expansion or compression of a perfect gas 
causes no change in its stock of internal energy since the temper- 
ature T is constant. During such an expansion the gas must 
take in an amount of heat just equal to the work it does, and 
conversely during an isothermal compression it must reject an 
amount of heat just equal to the work spent upon it. This 
quantity of heatQ (in B.t.u.) is, from equation (45), 

n wRT , V 2 t ,, 

Q = w ge v; (46) 

The following problem shows the application of the foregoing 
formulas to isothermal expansions: 

Air having a pressure of 100 pounds per square inch absolute 
and a volume of 1 cubic foot expands isothermally to a volume 
of 4 cubic feet. Find 

(a) External work of the expansion; 

(b) Heat required to produce the expansion; 

(c) Pressure at end of expansion. 

Solution, (a) Since the expansion is isothermal, 

External work, W = Pi7i log e * ^ 

= IOO X 144 X I x 1.3848 
= 19,940 foot-pounds. 

(b) Since the heat added equals increase in internal energy 
plus external work, and since the temperature remains constant 
(requiring therefore no heat to increase the internal energy), the 
internal energy equals zero and the heat added equals the work 
done. 

* 2.3 X log base 10 = log base e. Tables of natural logarithms are given in the 
appendix. 



32 



EXPANSION AND COMPRESSION OF GASES 



Then: Heat added = external work = 9>94Q = 25.6 B.t.u. 

778 



(c) 
then 



Since 



P1V1 

100 X 1 

P 2 



P 2 F 2 , 

P 2 x 4, 

25 pounds per square inch absolute. 



If a gas expands and does external work without receiving a 
supply of heat from an external source, it must derive the amount 
of heat needed to do the work from its own stock of internal 
energy. This process is then necessarily accompanied by a low- 
ering of temperature and the expansion obviously is not iso- 
thermal. 

4. Adiabatic Expansion and Compression. In the adiabatic 
mode of expansion or compression the working substance neither 
receives nor rejects heat as it expands or is compressed. A 
curve which shows the relation of pressures to volumes in such 
a process is called an adiabatic line (see Fig. 6). In any adiabatic 




Volume 
Fig. 6. — Isothermal and Adiabatic Expansion Lines. 

process the substance is neither gaining nor losing heat by 
conduction, radiation or internal chemical action. Hence the 
work which a gas does in such an expansion is all done at the 
expense of its stock of internal energy, and the work which is 
done upon a gas in such a compression all goes to increase its 
internal energy. Ideally adiabatic action could be secured by 
a gas expanding, or being compressed, in a cylinder which in all 
parts was a perfect non-conductor of heat. The compression of 
a gas in a cylinder is approximately adiabatic when the process is 
very rapidly performed, but when done so slowly that the heat 



ADIABATIC EXPANSION AND COMPRESSION 33 

has time to be dissipated by conduction the compression is more 
nearly isothermal. Fig. 6 shows on a pressure-volume diagram 
the relation between an isothermal and an adiabatic form of ex- 
pansion or compression from an initial condition P1V1 at A to 
final conditions at B and C for expansions, and at D and E for 
compressions. 

In order to derive the pressure-volume relation for a gas 
expanding adiabatically, consider the fundamental equation 
(page 20): 

Heat added = increase in internal energy + external work, or: 

.<2 = wK v (T 2 - TO + PdV (foot-pounds), (47) 

where K v is the specific heat in foot-pound units, i.e., 778 C v . 
In the adiabatic expansion no heat is added to or taken away 
from the gas by conduction or radiation, and, therefore, the left 
hand member of the above equation becomes zero. Furthermore, 
since equation (25) can always be applied to perfect gases, the 
following simultaneous equations may be written: 

o = wK v dT + P dV, (48) 

PV = wRT. (49) 

When P, V and T vary, as they do in adiabatic expansion, 
equation (49) may be written as follows: 

PdV+ VdP = wRdT, (50) 

and 

PdV + VdP 



dT = 



wR 



Substituting the value of dT in (48), 

n^T/ , v PdV + VdP , , 

PdV + wK v l — = o. (51) 

wR 

RPdV + K V P dV + K V V dP = o. 



34 EXPANSION AND COMPRESSION OF GASES 



To separate the variables divide by PV: 
W 4. K dV + K dF 



p dV dV dP . . 

R — + K, — + K, — = o. (52) 



Collecting terms, 

(i? + K.) ^ + tf. ^ = o. 
Integrating, 

(R + K v ) log e F + A% log e P = a constant = c. (53) 
*±*'log a F + log.P = C . 

Ay 

ig + gp 

lo ge PF * =c, (54) 

since from equation (^), 

R + K v = Kp y 

where K v and A% are respectively the specific heats at constant 
pressure and at constant volume in foot-pound units, then equa- 
tion (54) becomes: 



From equation (34), 



\og & PV Kv =c. 



therefore : 



r~ = y > or k~ = 7 > 



PV 7 = a constant. (55) 

Following the method used for obtaining an expression for 
the work done in isothermal expansion (equation 43), the work 
done, W (in foot-pounds), for a change of volume from Vi to F 2 , 

Jf*Vi 
PdV. (56) 



W 

V 2 



For purposes of integration, P can be substituted in terms of 
V as outlined below. In the general expression PV n = c,sl con- 
stant (see equation 35), where P and V are values of pressure and 
volume for any point on the curve of expansion of a gas of which 



ADIABATIC EXPANSION AND COMPRESSION 35 

the initial condition is given by the symbols Pi and V h we can 
then write, 

p = — (57) 

And substituting (57) in (56), 



r v~ n+1 Y 2 

L- n + ij 7l 

= 1 i-„ J (59) 

Since PV n = c = PiV? = P 2 V 2 n , we can substitute for c in 
(59) the values corresponding to the subscripts of V as follows: 

P 2 V 2 n V 2 l ~ n - PiF/% 1 -" 



W = 

w = 



1 — n 

P 2 V 2 - P 1 V 1 

) 

1 — n 



or W = ^ ^^ (foot-pounds). (60) 

n — 1 

Since PF = wPP, 

PT = wR (Tl ~ T2) - (foot-pounds). (61) 

n — 1 

Equations (60) and (61) apply to any gas undergoing expan- 
sion or compression according to PV n = a constant. In the case 
of an adiabatic expansion of a perfect gas n = 7 (see equation 

55). 
Change of Internal Energy During Adiabatic Processes. 

Since in an adiabatic expansion no heat is conducted to or away 
from the gas, the work is done at the expense of the internal 
energy and, therefore, the latter decreases by an amount equiva- 
lent to the amount of work performed. This loss in internal 
energy is readily computed by equation (60) or (61). The 
result must be divided by 778 in order to be in B.t.u. 
During an adiabatic compression the reverse occurs, i.e., there 



36 EXPANSION AND COMPRESSION OF GASES 

is a gain in internal energy and the same formulas apply, the 
result coming out negative, because work has been done on the 
gas. 

Relation between Volume, Pressure and Temperature in 
Adiabatic Expansion of a Perfect Gas. Since P, V and T vary 
during an adiabatic expansion, it will be necessary to develop for- 
mulas for obtaining these various quantities. It will be remem- 
bered that equation (25), applies to perfect gases at all times. 
Therefore, in the case of an adiabatic expansion or compression 
we can write the two simultaneous equations: 

■££-.££•• (A) 

±1 1 2 

p x v x y = p 2 v 2 y . (B) 

By means of these two equations we can find the final condi- 
tions of pressure, volume and temperature, having given two 
initial conditions and one final condition. 

For instance, having given V\, V 2 and T h to find T 2) divide (A ) 
by {B), member for member. Then 

V 1 V 2 



TxV^ T 2 V 2 y 
T 2 = Vj-v 

T 2 = T,' 



or 






In like manner the following formulas can be obtained: 

(63) 



p 



-rM y . («4) 



ADIABATIC EXPANSION AND COMPRESSION 37 

P* = Pi (ffi*. (65) 

V, = V 1 (^p, (66) 

1 

Fs=Fl (S) 71 ' (67) 

It should be noted that the above formulas can be used for 
any expansion of a perfect gas following PF" = a constant, 
provided y in the formulas is replaced by n. 

It is also to be noted that these equations can be used for 
any system of units so long as the same system of units is 
employed throughout an equation. 

There are many cases of expansions which are neither adiabatic 
nor isothermal and which are not straight lines on P-V diagrams. 
It will be observed from the equations in the discussion of the 
internal work done by an expanding gas and for the change of 
internal energy, that if in the general equation PV n = a. constant 
the exponent or index n is less than y, the work done is greater 
than the loss in internal energy. In other words for such a 
case, the expansion lies between an adiabatic and isothermal and 
the gas must be taking in heat as it expands. On the other 
hand, if n is greater than y the work done is less than the loss 
of internal energy. 

Example. Given a quantity of pure air in a cylinder at a tem- 
perature of 6o° F. (7\ = 460 + 60 = 520 degrees absolute) which 
is suddenly (adiabatically) compressed to half its original volume. 

Then — 1 = -— , and taking y as 1.405, the temperature immediately 
V 2 1 

after compression is completed, T 2 , is calculated by equation (62) 
as follows: 

/FA 7-1 /2V- 405 - 1 

T 2 = Ti I y) = 520 I- J = 520 X 2 0405 = 688° absolute, 

or t 2 in ordinary Fahrenheit is 688 — 460 or 228 degrees. 



38 EXPANSION AND COMPRESSION OF GASES 

The work done in adiabatic compression of cAe pound of this 
air is calculated by equation (61): 

w= wRjT.-T,) = 53.3(520-688) = 53-3 (-168) = _ 22jIIQ 
7-1 1.405 -1 0.405 

foot-pounds per pound of air compressed. The negative sign 
means that work has been done on the gas or the gas was com- 
pressed. If the sign had been positive it would have indicated 
an expansion. 

As the result of this compression the internal energy of the gas 

has been increased by — B.t.u., but if the cylinder is a con- 

_ 778 

ductor of heat, as in practice it always is, the whole of this heat 
will become dissipated in time by conduction to surrounding 
air and other bodies, and the internal energy will gradually 
return to its original value as the temperature of the gas comes 
back to the initial temperature of 6o° F. 

PROBLEMS 

1. Calculate the heat required to produce a temperature change of 50 F. 
in three pounds of air at constant pressure. 

2. How many foot-pounds of work are done by 2 lbs. of air in expand- 
ing to double its volume at a constant temperature of ioo° F.? 

3. Three pounds of air are to be compressed from a volume of 2 to 1 cu. ft. 
at a constant temperature of 6o° F. How many B.t.u. of heat must be 
rejected from the air? 

4. An air compressor has a cylinder volume of 2 cu. ft. If it takes air 
at 14.4 lbs. per sq. in. absolute and 70 F. and compresses it isothermally to 
100 lbs. per sq. in. absolute, find 

(a) Pounds of air in cylinder at beginning of compression stroke. 

(b) The final volume of the compressed air. 

(c) The foot-pounds of work done upon the gas during compres- 

sion. 

(d) The B.t.u. absorbed by the air in increasing the internal energy, 
(c) The B.t.u. to be abstracted from the cylinder. 

5. A cubic foot of air at a pressure of 150 lbs. per sq. in. gage expands 
isothermally until its pressure is 50 lbs per sq. in. gage. Calculate the 
work done during this expansion. 



PROBLEMS 39 

6. Air at ioo lbs per sq. in. absolute and a volume of 2 cu. ft. expands 
along an» = 1 curve to 25 lbs. per sq. in. absolute pressure. Find 

(a) Work done by the expansion. 

(b) Heat to be supplied. 

7. A quantity of air at 100 lbs. per sq. in. absolute pressure has a tem- 
perature of 8o° F. It expands isothermally to a pressure of 25 lbs. per 
sq. in. absolute when it has a volume of 4 cu. ft. Find (1) the mass of air 
present, (2) Work of the expansion in foot-pounds, (3) Heat required in 
B.t.u. 

8. Air at 100 lbs. per sq. in. absolute pressure and 2 cu. ft. expands to 
25 lbs. per sq. in. absolute adiabatically. What is the final volume? 

9. One cubic foot of air at 6o° F. and a pressure of 15 lbs. per sq. in. 
absolute is compressed without loss or addition of heat to 100 lbs. per sq. 
in. absolute pressure. Find the final temperature and volume. 

10. Two pounds of air are expanded from a temperature of 300 F. to 
200 F. adiabatically. How many foot-pounds of work are developed? 

11. A quantity of air having a volume of 1 cu. ft. at 6o° F. under a pres- 
sure of 100 lbs. per sq. in. absolute is expanded to 5 cu. ft. adiabatically^ 
Find the pounds of air present, the final temperature of the air and the 
work done during this expansion. 

12. Data the same as in Problem 4 but the compression is to be 
adiabatic. Find 

(a) The final volume of the compressed air. 

(b) The final temperature of the compressed air. 

(c) The foot-pounds of work required to compress this air. 

(d) The B.t.u. absorbed by the air in increasing the internal 

energy. 

(e) The B.t.u. to be abstracted from the gas. 

13. A pound of air at 32 F. and atmospheric pressure is compressed to 
4 atmospheres (absolute). What will be the final volume and the work 
of compression if the compression is (a) isothermal, (b) adiabatic? 

14. Plot the curve PV n = C, when n = 1.35. Initial pressure is 460 
lbs. per sq. in. gage, initial volume 0.5 cu. ft., and final volume 8 cu. ft. 

15. Prove that the work of an adiabatic expansion can be expressed by 
the formula: 

"—[-(in 



CHAPTER IV 

CYCLES OF HEAT ENGINES USING GAS 

The Heat Engine Cycle. In the heat engine, the working sub- 
stance or heat medium undergoes changes in its physical proper- 
ties converting heat into mechanical work. The series of such 
changes by the repetition of which the conversion of heat into 
work takes place forms the heat engine cycle.* The heat engine 
cycle usually consists of four events which are: heating, cooling, 
expansion, and compression. 

i. THE CARNOT CYCLE 

Very important conclusions regarding theoretically perfect 
heat engines are to be drawn from the consideration of the action 
of an ideal engine in which the working substance is a perfect 
gas which is made to go through a cycle of changes involving 
both isothermal and adiabatic expansions and compressions. 
This ideal cycle of operations was invented and first explained in 
1824 by Carnot, a French engineer. This cycle gave the first 
theoretical basis for comparing heat engines with an ideally 
perfect engine. The ideal Carnot cycle requires an engine, 
illustrated in Fig. 7, which consists of the following parts: 

(1) A piston and cylinder, as shown in Fig. 7, composed of 
perfectly non-conducting material, except the cylinder-head (left- 
hand end of the cylinder) which is a good conductor of heat. The 
space in the cylinder between the piston and the cylinder head is 
occupied by the working substance, which can be assumed to be 
a perfect gas. 

* A thermodynamic machine performing a cycle in which heat is changed into 
work is called a heat engine, and one performing a cycle in which heat is trans- 
ferred from a medium at a low temperature to one at a higher temperature is 
called a refrigerating machine. 

40 



THE CARNOT CYCLE 



41 



(2) A hot body H of unlimited heat capacity, always kept at a 
temperature TV. 

(3) A perfectly non-conducting cover N. 

(4) A refrigerating or cold body R of unlimited heat receiving 
capacity, which is kept at a constant temperature T% (lower 
than Ti). 



H 



N 




Fig. 7. — Apparatus and Diagram Illustrating a Reversible (Carnot) Cycle. 

It is arranged that H, N or R can be applied, as required, to 
the cylinder head. Assume that there is a charge of one pound 
of gas in the cylinder between the piston and the cylinder head, 
which at the beginning of the cycle, with the piston in the posi- 
tion shown, is at the temperature T h has a volume V a , and has a 
pressure P a . The subscripts attached to the letters V and P 
refer to points on the pressure- volume diagram shown in Fig. 7. 



42 CYCLES OF HEAT ENGINES USING GAS 

This diagram shows, by curves connecting the points a, b, c and 
d, the four steps in the cycle. 

The operation of this cycle will be described in four parts as 
follows: 

(i) Apply the hot body or heater H to the cylinder head at 
the left-hand side of the figure. The addition of heat to the gas 
will cause it to expand isothermally along the curve ab, because 
the temperature will be maintained constant during the process 
at T\. The pressure drops slightly to P b when the volume be- 
comes Vb. During this expansion external work has been done 
in advancing the piston and the heat equivalent of this work has 
been obtained from the hot body H. 

(2) Take away the hot body H and at the same time attach 
to the cylinder head the non-conducting cover N. During this 
time the piston has continued to advance toward the right, 
doing work without receiving any heat from an external source, 
so that the expansion of the gas in this step has been done at 
the expense of the stock of internal energy in the gas along the 
adiabatic curve be. The temperature has continued to drop in 
proportion to the loss of heat to the value T 2 . Pressure is then 
P c and the volume is V c . 

(3) Take away the non-conductor N and apply the refriger- 
ator R. Then force the piston back into the cylinder. The 
gas will be compressed isothermally at the temperature T 2 . 
In this compression, work is being done on the gas, and heat is 
developed, but all of it goes into the refrigerator R, in which the 
temperature is always maintained constant at T 2 . This com- 
pression is continued up to a point d in the diagram, so selected 
that a further compression (adiabatic) in the next (fourth) 
stage will cause the volume, pressure and temperature to reach 
their initial values as at the beginning of the cycle* 

(4) Take away the refrigerator R and apply the non-conduct- 
ing cover N. Then continue the compression of the gas without 

* Briefly the third stage of the cycle must be stopped when a point d is reached, 
so located that an adiabatic curve (PV 7 = constant) drawn from it will pass 
through the "initial" point a. 



THE CARNOT CYCLE 43 

the addition of any heat. It will be the adiabatic curve da. 
The pressure and the temperature will rise and, if the point d has 
been properly selected, when the pressure has been brought back 
to its initial value P a the temperature will also have risen to its 
initial value 2\. The cycle is thus finished and the gas is ready 
for a repetition of the same series of processes comprising the 
cycle. 

To define the Carnot cycle completely we must determine 
how to locate algebraically the proper place to stop the third 
step (the location of d). During the second step (adiabatic ex- 
pansion from b to c) by applying equation (62), the following 
temperature and volume relations exist: 

Ti [Vol 7 - 1 
T 2 lV b ] ' 

also for the adiabatic compression the fourth step can be sirni- 
lary stated, 

Hence, 



wr - wr 



Simplifying and transposing, we have 

V b V c 



v. v, < 68) 



V 



— is the ratio of expansion r for the isothermal expansion in 

V a 

the first step of the cycle. This has been shown (Equation 68) to 

be equal to — - in the isothermal compression in the third step in 

order that the adiabatic compression occurring in the fourth 
step shall complete the cycle. 

A summary of the heat changes to and from the working 
gas (per pound) in the four steps of the Carnot cycle is as fol- 
lows: 



44 CYCLES OF HEAT ENGINES USING GAS 

(ab). Heat taken in from hot body (by equation (45), in 
foot-pounds) is: 

RTl\0g e ^- (69) 

• V a 

(be). No heat taken in or rejected. 

(cd). Heat rejected to refrigerator (by equation (45), in 
foot-pounds) is: 

V V * 

RT 2 log, — = - RT 2 log e -/• (70) 

V c V d 

(da). No heat taken in or rejected. 
Hence, the net amount of work done, W, by the gas in this 
cycle, being the mechanical equivalent (foot-pounds) of the 
excess of heat taken in over that rejected, is the algebraic sum 
of (69) and (70): 

W = r(t 1 log, y a -T, log, £) =R(T l - r 2 ) loge £• (71) 

The thermal or heat efficiency of a cycle is defined as the ratio of 

Heat equivalent of work done 
Heat taken in 

The heat equivalent of work done is, by equation (71), 

ic^-T^log,^, 

and the heat taken in is, by equation (69), 
The ratio above representing the efficiency E is: 

r (z\ - r 2 ) log. ^ 

E ^ = T -^- (72) 

RT, log, £ 

V a 

Log ---togpj. 



REVERSIBLE CYCLES 45 

The efficiency of the Carnot cycle as expressed by equation (72) 
is the maximum possible theoretical efficiency which may be 
obtained with any heat engine working between the temperature 
limits 7\ and T 2 . This equation can be used as a standard for the 
comparison of the efficiencies of actual heat engines working 
between two temperature limits. 

Reversible Cycles. A heat engine which is capable of dis- 
charging to the "source of heat" when running in the reverse 
direction from that of its normal cycle the same quantity of 
heat that it would take from this source when it is running 
direct and doing work is said to operate with its cycle reversed, 
or, in other words, the engine is reversible. A reversible heat 
engine then is one which, if made to follow its indicator diagram 
in the reverse direction, will require the same horse power to drive 
it as a refrigerating machine as the engine will deliver when 
running direct, assuming that the quantity of heat used is the 
same in the two cases. An engine following Carnot's cycle is, 
for example, a reversible engine. The thermodynamic idea of 
reversibility in engines is of very great value because no heat 
engine can be more efficient than a reversible engine when both 
work between the same limits of temperature; that is, when 
both engines take in the same amount of heat at the same higher 
temperature and reject the same amount at the same lower 
temperature. 

It was first proved conclusively by Carnot that no other heat 
engine can be more efficient than a reversible engine when both 
work between the same temperature limits. To illustrate this 
principle, assume that there are two engines A and B. Of these 
let us say A is reversible and B is not. In their operation both 
take heat from a hot body or heater H and reject heat to a refrig- 
erator or cold body R. Let Q H be the quantity of heat which 
the reversible engine A takes in from the hot body H for each 
unit of work performed, and let Q B be the quantity of heat per 
unit of work which it discharges to the refrigerator R. 

For the purpose of' this discussion, assume that the non- 
reversible engine B is more efficient than the reversible engine A. 



46 CYCLES OF HEAT ENGINES USING GAS 

Under these circumstances it is obvious that the engine B will take 
in less heat than A and it will reject correspondingly less heat 
to R per unit of work performed. The heat taken in by the 
non-reversible engine B from the hot body H we shall designate 
then by a quantity less than Q H , or Q H — X, and the heat rejected 
by B to the refrigerator R by Q R — X. Now if the non-reversible 
engine B is working direct (when converting heat into work) 
and is made to drive the reversible engine A according to its 
reverse cycle (when converting work into heat), then for every 
unit of work done by the engine B in driving the reversible 
engine A, the quantity of heat mentioned above, that is, Q H — X, 
would be taken from the hot body H by the non-reversible 
engine B and, similarly, the quantity of heat represented by Q H 
would be returned to the hot body H by the reverse action of 
the cycle of operations performed by A. This follows because 
the engine A is reversible and it returns, therefore, to H, when 
operating on the reverse cycle, the same amount of heat as it 
would take in from H when working on its direct cycle. By 
this arrangement the hot body H would be continually receiving 
heat, in the amount represented by X for each unit of work 
performed. At the same time the non-reversible engine B dis- 
charges to the refrigerator R a quantity of heat represented by 
Q R — X, while the reversible engine A removes from the refriger- 
ator R a quantity represented by Q R . As a result of this last 
operation the cold body will be losing continually per unit of work 
performed a quantity of heat equal to X. The combined per- 
formances of the two engines, one working direct as a normal heat 
engine and the other, according to its reverse cycle, as a com- 
pressor or what might be called a "heat pump/' gives a constant 
removal of heat from the refrigerator R to the hot body H, 
and as a result a degree of infinite coldness must be finally pro- 
duced in the refrigerator. 

If we assume that there is no mechanical friction, this com- 
bined machine, consisting of a normal heat engine and com- 
pressor, will require no power from outside the system. For 
this reason the assumption that the non-reversible engine B 



REVERSIBLE CYCLES 



47 



can be more efficient than the reversible engine A has brought 
us to a result which is impossible from the standpoint of expe- 
rience as embodied in the statement of the " Second Law of 
Thermodynamics "; that is, it is impossible to have a self-acting 
engine capable of transferring heat, infinite in quantity, from a 
cold body to a hot body. We should, therefore, conclude that 
no non-reversible engine, as B for example, can be more efficient 
than a reversible engine A when both engines operate between 
the same temperature limits. More briefly, when the source of 
heat and the cold receiver are the same for both a reversible heat 
engine and any other engine, then the reversible engine must 
have a higher possible efficiency; and if both engines are reversible 
it follows that neither can be more efficient than the other. 

A reversible engine is perfect from the viewpoint of efficiency; 
that is, its efficiency is the best obtainable. No other engine 
than a reversible engine which takes in and discharges heat at 
identical temperatures will transform into work a greater part of 
the heat which it takes in. Finally, it should be stated as regards 
this efficiency that the nature of the substance being expanded or 
compressed has absolutely no relation to the thermal efficiency 
as outlined above. 

If an engine operating on Carnot's cycle is reversed in its action, 
so that the same indicator diagram shown in Fig. 7 would be 
traced in the opposite direction, the reversed cycle, when begin- 
ning as before at point a with a perfect gas at the temperature 
Ti, will consist of the following stages: 

(1) When the non-conductor N is applied and the piston is 
advanced toward the right by the source of power performing 
the reversed cycle, the gas will expand, tracing the adiabatic 
curve ad, with constant lowering of temperature which at the 
point d will be T 2 . 

(2) When the non-conductor N is now removed, the refrig- 
erator R is applied, and the piston continues on its outward stroke. 
The gas will expand isothermally at the constant temperature T 2 , 
tracing the curve dc. During this stage the gas is taking heat 
from the refrigerator R. 



48 CYCLES OF HEAT ENGINES USING GAS 

(3) When the refrigerator R is removed and the non-conductor 
N is again applied, which will be on the back stroke of the engine, 
the gas will be compressed, and on the indicator diagram another 
adiabatic curve cb will be traced. At the point b the temper- 
ature will be obviously T\. 

(4) When the non-conductor N is removed and the hot body 
H is again applied, with the compression continuing along the 
isothermal curve ba, heat will be discharged to the hot body H, 
while the temperature is maintained constant at 7\. 

The cycle has now been traced in a reverse direction from the 
beginning back to the starting point at a, and is now complete. 
During this process an amount of work represented by the area 
of the indicator diagram, equivalent in foot-pounds to 

R log e — - (7\ — T 2 ) (see equation 71), 

V a 

has been converted into heat. First, heat was taken from the 
refrigerator R, represented in amount by 

RT 2 logo ■=£■> 

V d 

and second, heat was rejected to the hot body H in the amount 
RTylog^ or -2?7\log3. 

As in direct operation of Carnot's cycle no heat is given or lost 
in the first and third stages outlined above. The algebraic sum 

of these two quantities, remembering that — = -=^- } gives the 

V a V d 

net amount of work done, W, on the gas, and, therefore, the 
net amount of heat (foot-pound units) transferred from the cold 
body R to the hot body H or, 

W = RT 2 log e ^ - RT, log, £ = -R log c £ (7\ - T 2 ). (73) 

V a y a y a 



HOT-AIR ENGINE CYCLES 49 

Since the result is the same as given by equation (72), although 
opposite in sign on account of being work of compression, it 
will be observed that in the reverse cycle the same amount 
of heat is given to the hot body H as was taken from it in the 
direct operation of the same cycle, and that the same amount 
of heat is now taken from the refrigerator R as was in the other 
case given to it. 

2. HOT-AIR ENGINE CYCLES 

The hot-air engine is a non-explosive type of external combus- 
tion heat engine, the working substance being atmospheric air 
which undergoes no change in its physical state. This type of 
prime-mover was invented about 100 years ago. It is little used 
on account of the difficulty in transmitting heat through the 
metallic walls to the dry gas air. Then the hot-air engine is 
very bulky in proportion to its power. There are also difficul- 
ties in carrying out practically the theoretical cycles of operation, 
on account of the rapid deterioration of the heat-conducting sur- 
faces. The mechanical efficiency of the hot-air engine is also low. 

The advantages of the hot-air engine are ease of operation and 
safety. Also its speed being low, on account of the rate of heat 
transmission, it is well suited for the driving of small pumps and 
for other domestic uses where small powers are required and also 
where the fuel-economy is a matter of minor importance. 

Hot-air engine cycles are divided into two groups : — 

Group I. External-combustion hot-air engines with a closed 
cycle and constant- volume temperature changes. 

Group II. External-combustion hot-air engines with an open 
cycle and constant-pressure temperature changes. 

In the hot-air engines an attempt was made to put in practice 
a cycle of the ideal Carnot type with the addition of a regenera- 
tor. The regenerator was a storage-battery of heat, its function 
being to absorb, store and return heat rapidly, replacing the 
adiabatic expansion curves of the Carnot cycle by lines of con- 
stant volume in group I and by lines of constant pressure in 



5o 



CYCLES OF HEAT ENGINES USING GAS 



group II. The regenerator consists of a chamber containing 
strips of metal, coils of wire, or any other heat-absorbing material 
arranged in such a manner as to present a very large surface to 
the air passing through it. 

The Stirling Engine Cycle. The Stirling engine belongs to 
group I, the regenerator being so arranged that the pressure 
chops with the temperature at such a rate as to keep the vol- 
ume constant. It is an external-combustion engine and its 



I 


) 






\> 




V*- 




N& 




^&> 




^ 









a 


v> 




X r 4„ 










d 



Volume 
Fig. 8. — Stirling Engine Cycle. 

cycle of operation is closed, the same air is used over and 
over again, any loss by leakage being supplied by a small 
force-pump. The engine consists, essentially, of a pair of dis- 
placer cylinders, a double-acting working cylinder, a regenera- 
tor and a refrigerator. In the earlier forms the displacer cyl- 
inder had a double wall, the regenerator and refrigerator being 
placed in the annular space surrounding the displacer cylinder. 
In the latter forms the regenerator and refrigerator are arranged 
separately from the displacer cylinder but in direct communi- 
cation with its top and bottom. A plunger works in the dis- 
placer cylinder, this plunger being filled with a non-conducting 
material such as brick-dust. The engine derives its heat by 



STIRLING ENGINE CYCLE 5 1 

conduction from a furnace which is placed beneath the displac- 
ing cylinder. 

In the cycle represented by Fig. 8, the action of the engine 
is carried out as follows: 

1. The substance, air, having a volume V a , a pressure P a and 
a temperature T a receives heat at constant volume by passing 
through the regenerator, and receives also heat from the furnace. 
As a result of the heat addition, its pressure is increased to P b , its 
temperature to T b , this process being represented by the constant 
volume line ab, Fig. 8. During this event the plunger is moving up. 

2. The pressure under the working piston increases as a result 
of the addition of heat and the expansion of the air follows, 
producing the working stroke. This expansion is represented by 
be and is an isothermal, the air receiving heat from the furnace 
plates and the temperature remaining 7\. 

3. The plunger descends, displacing the hot air and forcing the 
same through the regenerator and refrigerator. As a result of 
this, the temperature drops from T c to T& and the pressure is 
decreased from P c to P d , the volume remaining constant as shown 
by the line cd. 

4. Due to this cooling effect the working piston descends com- 
pressing the air. This compression, represented by da, is iso- 
thermal, the air being cooled during the process by the regenera- 
tor and refrigerator to remain at the temperature TV 

Referring to Fig. 8 of the Stirling cycle, the useful work W 
is in foot-pounds, 

W = (jP b V b log, y>) - (p d V d log, y)' (H) 

The mean effective pressure is in pounds per square foot, 

M.E.P. = W ■ (75) 

V d — V a 

The horse power developed can be represented by 

H.P. = 5^- ( 7 6) 

33,000 

where N = revolutions (cycles) per minute. 



52 



CYCLES OF HEAT ENGINES USING GAS 



Since the expansion and compression processes are isothermal, 
the efficiency of the Stirling cycle is: 



P b V b log e ^-P d V d log ] 



E 



Vt 



V a 



P b V b log e 



Vc 
V b 



(77) 



Since V a = V b and V c = V d , also RTi = P b V b and RT 2 = 
PdV d , equation (77) becomes, 



E = 



RTr loge — c - RT^loge -=/ 

Vb V b 

Vc 

RT ± loge ^r 

Vb 



Tt -T 2 



(78) 



The Ericsson Engine Cycle. This engine belongs to group II, 
the regenerator changing the temperature at constant pressure. 
This engine consists of five parts : a compressing pump, a receiver, 
a regenerator, a refrigerator and a working cylinder. The ideal 



a' 

6 

m 

CL, 



a 


b 


\Ti 


• 




NO* r ^ 

\Isothermal 


^^sothermal • 

-r 




d 





Volume 
Fig. 9. — The Ericsson Engine Cycle. 

cycle for this engine is represented by Fig. 9, the order of events 
being as follows : 

Atmospheric air is drawn into the compressing pump as shown 
by d'd. This air is compressed isothermally to a, during the 
return stroke of the pump; it is then forced into a receiver as 
shown by aa! . Thus d'daa' represents the pump cycle. 

As the working cylinder begins its forward or up stroke, the 



THE LENOIR ENGINE 53 

compressed air is admitted into the working cylinder at constant 
pressure. As this admission from a to b is through the regenera- 
tor, the entering air takes up heat increasing in temperature from 
Ti to T\. As soon as the supply of compressed air is cut off, 
the air expands isothermally along be to atmospheric pressure, 
while heat is being supplied by a furnace at the bottom of the 
working cylinder. During the return stroke of the working pis- 
ton, the air is discharged at constant pressure through the regen- 
erator, giving up its heat and is cooled to the temperature T 2 . 
In Fig. 9, a' bed' is the cycle of the working cylinder, the net work 
being represented by abed. 

The Ericsson cycle has the same efficiency as the Stirling cycle, 
the regenerator process being carried out at constant pressure 
instead of at constant volume. 

3. INTERNAL-COMBUSTION ENGINE CYCLES 

The internal-combustion engine utilizes as its working sub- 
stance a mixture of air and gas or air and petroleum vapor. 
Combustion of the mixture takes place inside the engine cylinder, 
or in a communicating vessel, and the heat generated is converted 
into work. As the specific heat and the specific volume of the 
mixture do not differ much from that of air, the cyclical analy- 
sis and the theory of the internal-combustion engine are developed 
on the assumption that the working substance is air. The 
internal-combustion engine cycles can be divided into the follow- 
ing groups: 

Group I. Engines without compression. 

Group II. Compression engines. 

Group I. Internal-Combustion Engines Without Compression 

The Lenoir Engine. An engine of this group, invented by 
Pierre Lenoir in i860, was the first successful practical gas 
engine. The action of this type of engine is as follows : 

As the piston leaves the dead center it draws into the cylinder 
a charge of gas and air which is proportioned at the admission 



54 



CYCLES OF HEAT ENGINES USING GAS 



valve to form an explosive mixture. Admission is cut off some- 
what before half-stroke and this is followed by the ignition of the 
mixture by means of an electric spark. The explosion produces 
a rapid rise in pressure above atmospheric, thus forcing the piston 
to the end of the stroke. The exhaust valve opens near the end 
of the stroke and the burnt products are expelled during the return 
stroke of the piston. A fly-wheel carries the piston over during 
the exhaust stroke as well as also, during the suction stroke. 
The above series of operations takes place at each end of the 
piston, producing two impulses for each revolution. 

Fig. 10 represents the cycle of operations of the Lenoir engine 




Volume 
Fig. io. — Ideal Lenoir Cycle. 



on P-V coordinates. The drawing in of the mixture of gas and 
air at atmospheric pressure is represented by ab. Ignition takes 
place at b and is followed by the constant volume combustion 
line be. The working stroke, an adiabatic expansion, takes place 
to atmospheric pressure, as shown by cd, and the products of 
combustion are rejected during the return stroke of the piston da. 
Calling P, V, T the absolute pressure in pounds per square foot, 
the specific volume in cubic feet per pound of mixture, and the 
absolute temperature in degrees Fahrenheit respectively; also 
using subscripts b, c, d to designate the points at the corners bed 
of Fig. io, the following expressions will be obtained: 



THE LENOIR ENGINE 55 

If the heat developed by the complete combustion of the 
mixture be designated by Q h then 

6i = wC v (T c -T b ). (79) 

Since the combustion of the mixture takes place at constant 
volume, 

V c = V b 

T c = T b + 2*. (80) 



For unit weight, 



p. = p> £• (81) 

l 6 



1 



(82) 



The expansion cd being adiabatic (n = 7), 

If Q 2 is the heat rejected by the engine after performing its 
cycle of operations, 

Q 2 = wC v (T d - T b ). .(84) 

The useful work available is : 

W = J(Q 1 - Q 2 ). (85) 

The mean effective pressure is the average unbalanced pressure 
on the piston of the engine in pounds per unit area, and using 
symbols and units as on page 51, 

M.E.P. = ir i V- (S6) 

V d — V b 

The horse power developed would be found by equation 

h.p. = K2S2L, (87) 

33,000 

where N designates the number of explosions per minute. 
The cycle efficiency for the conditions of the problem is: 

7? _ Qi ~ & ft _ T d — T b , RR x 

E -^or- I -Q 1 - t T zwr (88) 



56 



CYCLES OF HEAT ENGINES USING GAS 



Group II. Compression Engines 

The first compression engine was patented as early as 1799 by 
Philip Lebon. The thermodynamic advantages of compression 
before ignition will be evident from the demonstrations which 
follow. A mechanical advantage due to compression is the re- 
duced shock due to the explosion and the resulting improved, 
balance of parts. 

1. The Otto Cycle. The Otto cycle, embodying the principles 
first proposed by Beau de Rochas in 1862, resulted in the most 
successful internal-combustion engine of to-day. This cycle was 
adopted by Dr. Otto in 1876, the operations being carried out in 
four consecutive equal strokes of the piston, requiring two com- 
plete revolutions of the engine crank-shaft. The ideal diagram 




Volume 
Fig. 11. — Otto Cycle. 

for the Otto cycle takes the form shown by Fig. 11, the operation 
being as follows: 

A mixture of gas and air is drawn in during the complete for- 
ward stroke of the piston, as shown by a' a. The return of the 
piston compresses the mixture along the adiabatic curve ab. 
Explosion of the compressed charge takes place at b, with the 
consequent combustion at constant volume to c. cd is the adia- 



THE BRAYTON CYCLE 57 

batic expansion producing the second forward stroke. The ex- 
haust valve opens at d, cooling the gases to the exhaust pressure a, 
and rejecting them to the atmosphere. 
The heat added during the combustion from b to c is 

<2i = wC, (T c - T b ). (89) 

The heat rejected from d to a is 

ft = wC, (T d - T a ). (90) 

The cycle efficiency is 

E _ Qi-Q2 _ wC v {T c - T b ) -wC, {T d - T a ) _ Tg-Tg , . 
Qx wC v {T c -T b ) J T c -T b ' l9Ij 

Since the expansion and compression are adiabatic the follow- 
ing relations will hold: 

T a Va n - 1 = T b V b n ~\ and T c V c n - 1 = TaV^ 1 



d 



Since V c = V b and V a = V d , 

d -L c . I J- d J- a J- a 

7JT — 7fT> 2I S0 — — — — • 

J- a J- b -L c — J- & i&. 

Smce Tr(vj -Ur 

„ T a /VA"- 1 /PA"— i \ 

E = I -Tr I -{va) =i -{pJ n - (92) 

Equation (92) shows that the efficiency of the Otto cycle de- 
pends on the amount of compression before explosion. 

2. The Brayton Cycle. This cycle is often called the Joule 
cycle after its inventor, or the Brayton cycle after George B. 
Brayton, who in 1872 designed an engine with gradual constant 
pressure combustion. In this engine a mixture of gas and air 
is first compressed in a separate pump and forced into a receiver. 
On the way from the receiver to the engine cylinder the mixture is 
ignited by a gas jet which burns steadily without sudden expkx 
sion, producing temperature and volume changes at constant 
pressure. After expansion the piston drives out the products 
of combustion at atmospheric pressure. 

The cycle of operations for the Brayton engine is represented 



58 



CYCLES OF HEAT ENGINES USING GAS 



by Fig. 12. a' a represents the supply of the combustible mixture 
to the pump where it is compressed adiabatically to b and forced 
into a receiver, be represents the burning of the compressed 



6'- 



a'- 





b 


c 






- 




VAdiabatic 


\~~Adiabatic 


^d 






a 



Volume 
Fig. 12. — Bray ton Cycle. 

mixture at constant pressure. As the mixture enters the working 
cylinder, it expands adiabatically along cd. This is followed by 
the rejection of the burnt gases along the atmospheric line da. 
The heat added during the constant pressure combustion from 

b to c is 

Qi = wC P (Tc - T b ). (93) 

The heat rejected from d to a is 

Q 2 = wC P (T d — T a ). (94) 



E 



The cycle efficiency is 

= Qi — Q2 = wCp (Tc — Tb) —wC p (Tg — Tg) _ Td — Tg , , 
Qi wC v (T c -Tb) _I T c -Tb {95) 



Since the expansion and compression phases of the cycle are 
adiabatic and P c = Pb\ Pa = Pa, therefore, 

Td _ Tc 

Tg Tb 

J- d -La -L a 

Tc-Tb ~ Y b " X 



and 



or =■-©" « 



Equation (96) is the same as equation (92) and shows that the 
cycle efficiencies of the Otto and Br ay ton cycles are the same, 



THE DIESEL CYCLE 



59 



under the same initial conditions, and depend on the amount of 
compression of the charge before explosion. 

3. The Diesel Cycle. This cycle is carried out in four strokes 
of equal length, as in the case of the Otto cycle. Atmospheric 
air is drawn in during the complete forward stroke of the piston 
as shown by la (Fig. 13). The return of the piston compresses 




Fig. 13. — Diesel Engine Cycle. 

the air adiabatically to b, a pressure of about 500 pounds per 
square inch. At the end of the compression stroke a charge of 
the liquid fuel is injected in a finely divided form by an auxiliary 
pump or compressor and burns nearly at constant pressure when 
it comes in contact with the highly compressed air: The supply 
and combustion of the fuel, as represented by the combustion 
line be, is cut off at one-tenth to one-sixth the working stroke, 
depending upon the load. Expansion takes place during the 
balance of the stroke as shown by the adiabatic expansion line cd. 
Release occurs at d with the consequent drop in pressure, and the 
burnt gases are rejected during the fourth stroke of the cycle. 
The heat added (Qi) during the combustion of the fuel is: 

Q x = wC P (T c - T b ). (97) 

The heat rejected from d to a is 

Q 2 = wC v (T d - T a ). (98) 

The cycle efficiency is then, 



60 CYCLES OF HEAT ENGINES USING GAS 

E „ ft -Q2 „ wC P (Tc. - T b ) - wC v (T d - T a ) 
Qi ' wC P (T c - T b ) 

C v (T d — T a \ I (T d — T a \ / x 

" X -C, Kf^Yj = '-yKf^Tj (99) 

PROBLEMS 

i. A Carnot engine containing 10 lbs. of air has at the beginning of the 
expansion stroke a volume of 10 cu. ft. and a pressure of 200 lbs. per sq. in. 
absolute. The exhaust temperature is o° F. If 10 B.t.u. of heat are added 
to the cycle, find 

(a) Efficiency of the cycle. 

(b) Work of the cycle. 

2. A Carnot cycle has at the beginning of the expansion stroke a pressure 
of 75 lbs. per sq. in. absolute, a volume of 2 cu. ft. and a temperature of 
200 F. The volume at the end of the isothermal expansion is 4 cu. ft. 
The exhaust temperature is $0° F. Find 

(a) Heat added to cycle. 

(b) Efficiency of cycle. 

(c) Work of cycle. 

3. A cycle made up of two isothermal and two adiabatic curves has a 
pressure of 100 lbs. per sq. in. absolute and a volume of 1 cu. ft. at the begin- 
ning of the isothermal expansion. At the end of the adiabatic expansion 
the pressure is 10 lbs. per sq. in. absolute and the volume is 8 cu. ft. Find 

(a) Efficiency of cycle. 

(b) Heat added to cycle. 

(c) Net work of cycle. 

4. In a Carnot cycle the heat is added at a temperature of 400 F. and 
rejected at 70 F. The working substance is 1 lb. of air which has a volume 
of 2 cu. ft. at the beginning and a volume of 4 cu. ft. at the end of the 
isothermal expansion. Find 

(a) Volume at end of isothermal compression. 

(b) Heat added to the cycle. 

(c) Heat rejected from cycle. 

(d) Net work of the cycle. 

5. Air at a pressure of 100 lbs. per sq. in. absolute, having a volume of 
1 cu. ft. and a temperature of 200 F., passes through the following opera- 
tions: 

1st. Heat is supplied to the gas while expansion takes place under 
constant pressure until the volume equals 2 cu. ft. 



PROBLEMS 6l 

2d. It then expands adiabatically to 15 lbs. per sq. in. absolute 
pressure. 

3d. Heat is then rejected while compression takes place at con- 
stant pressure. 

4th. The gas is then compressed adiabatically to its original volume 
of 1 cu. ft. 
Find (a) Pounds of air used. 

(b) Temperature at end of constant pressure expansion. 

(c) Heat added to the cycle. 

(d) Net work of cycle. 

(e) Efficiency of cycle. 

6. A Stirling hot-air engine having a piston 16 inches in diameter and 
a stroke of 4 ft. makes 28 r.p.m. The upper temperature is 650 F., while 
the lower temperature is 150 F. Assuming that the volume of the working 
cylinder is one-half that of the displacer cylinder, calculate the pressures and 
volumes at each point of the cycle (Fig. 8) ; also calculate the mean effective 
pressure, the horse power developed, and the cycle efficiency. 

7. Plot the Stirling engine cycle from the results of problem 6. 

8. Compare the cycle efficiencies of internal-combustion engines working 
on the Otto cycle using the fuels and compression pressures as indicated 
below: 

Gasoline, 75; producer gas, 150; and blast furnace gas, 200 lbs. per sq. in. 
gage pressure. 

9. Calculate the theoretical pressures, volumes and temperatures at each 
point of an Otto cycle, as well as the mean effective pressure, horse power, 
and cycle efficiency for the following conditions of Fig. 11. Assume that 
the pressure after compression is 180 lbs. per sq. in. gage, that 80 B.t.u. 
are added during combustion and that n in PV n = 1.4 : 

P a = 14.7 lbs. per sq. in., V a = 13.5 cu. ft., 
T a = 7° + 459-5 = 5 2 9-5° F. absolute. 

10. Prove that cycle efficiency of the Diesel engine depends not only 
upon the compression pressure before ignition but also upon the point of 
cut-off C in Fig. 13. 

11. In what respects do the actual cycles of internal-combustion engines 
differ from the theoretical Otto and Diesel cycles? 

12. The demonstrations in the text for the various gas cycles were based 
on the assumption that the specific heats of gases are constant. Since the 
specific heats of gases vary at high temperatures, show the effect of this 
variability on the cyclic analysis. Refer to Lucke's Engineering Thermo- 
dynamics, Levin's Modern Gas Engines and the Gas Producer, and Clerk's 
Gas, Petrol and Oil Engines, Vol. II. 



CHAPTER V 
PROPERTIES OF VAPORS 

Saturated and Superheated Vapors. As was explained in the 
chapter on the properties of perfect gases, a vapor can be liqui- 
fied by pressure or temperature changes alone. At every pres- 
sure there is a fixed point, called the point of vaporization, at 
which a liquid can be changed into a vapor by the addition of 
heat. A vapor near the point of vaporization is called a saturated 
vapor and has a definite vaporization temperature for any given 
pressure. When a vapor is heated so that its temperature is 
greater than the vaporization temperature corresponding to a 
given pressure, it is said to be a superheated vapor. Superheated 
vapors only when far removed from the vaporization tempera- 
ture approach nearly the laws of perfect gases. 

Theory of Vaporization. When heat is added to a liquid its 
temperature will rise with a slight volume increase until the point 
of vaporization is reached. This is always a definite point for 
any given pressure and depends on the character of the liquid. 
Thus the point of vaporization of water at the atmospheric pres- 
sure of 14.7 pounds per square inch is 21 2 F., while that at a pres- 
sure of 150 pounds absolute is 358 F. On the other hand, the 
vaporization temperature of ammonia vapor at a pressure of 1 50 
pounds absolute is 79 F. The increased temperature of vapori- 
zation with the pressure increase is due to the fact that the mole- 
cules being crowded, the velocity per molecule, or temperature, 
must be great enough to overcome not only molecular attraction 
but also external pressure before vaporization can take place. 

When the point of vaporization is reached, any further addition 

of heat will not cause any temperature increase, but internal 

work accompanied by vaporization will be produced as well as 

an enormous increase in volume. The heat required to entirely 

62 



VAPOR TABLES 63 

vaporize a unit weight of a substance at a given pressure is termed 
the latent heat of vaporization. 

The volume developed when one unit weight of the liquid has 
been completely evaporated is called the specific volume of the 
dry vapor, this volume depending on the pressure. 

When vaporization is incomplete the vapor is termed a wet 
saturated vapor, which means that the vapor is in contact with 
the liquid from which it is formed. The percentage dryness in 
the vapor is called its quality. Thus the quality of steam is 0.97 
when one pound of it consists of 97 per cent steam and 3 per cent 
water. While the volume of a vapor increases with the quality, 
the temperature is constant between the point of vaporization 
and the point of superheat for any given pressure. Thus the 
temperature of steam vapor corresponding to 150 pounds abso- 
lute is 358 F., no matter whether the quality is 0.10, 0.75 or 1.00. 

Further addition of heat after complete vaporization will cause 
temperature as well as volume changes, the substance being in 
the superheated vapor condition. 

Vapor Tables. The exact quantities of heat required to pro- 
duce the above effects under various conditions, as well as the 

100 



80 



60 



40 



20 











M = 


Marl 


1 
s and Da% 


is 
































G = 
P = 


Good 
Peal) 


enou 
ody 


ffh 






























































M = 302.S 
G = 302.9 
P = 302.95 




























( 


( M = 

h= 


292.7 
292.7 
292.7 


Ui^. 


























<\ 


M = 267.3 
G= 267.2 . 
P = 267.26 ( 


IM = 281.0 
'<G =281.0 
i P=281.0 


i^*"" 


























' 


M = 250.3 V 
G =250.3 
P=250.34 






























( 


'M = 228.0 
G = 228.0 
P=327.95 






























f M = 
i G p = 


212° 
212° 
211.9 


!6 


\ 



















































































































240 260 280 

Temperature, Degrees Fahrenheit 



220 
Fig. 14. — Pressure-Temperature Relations for Saturated Steam 



300 



relations existing between pressure, volume, and temperature of 
saturated and superheated vapors have been determined exper- 



6 4 



PROPERTIES OF VAPORS 



imentally. These experimental results have been given in the 
form of empirical equations from which vapor tables have been 
computed. Tables 3 and 4, showing the properties of dry satu- 




100 



•3 80 

TIX 

-< 

a 
•- 60 

a* 



40 



20 



12 14 16 18 
Volume, cu. ft. per lb. 



Fig. 15. — Pressure- Volume Relations for Saturated Steam. 





























M = Mark 
G = Good 
P = Peat 


s and Da 1 

enough 

ody 


as 








P^ ' 







































































880 



Fig. 16. 



900 920 940 960 

Latent Heat 

Pressure-Latent Heat Relations for Saturated Steam. 



rated steam and of ammonia, are given in the appendix. This 
text should be supplemented by the more complete tables of 
Marks and Davis, Peabody or Goodenough. 



TEMPERATURE, PRESSURE AND VOLUME OF STEAM 65 

In Figs. 14, 15 and 16 are plotted the several variables, as given 
in the above-mentioned tables, for various pressures. 

In most vapor tables will be found, corresponding to the pres- 
sure of the vapor in pounds per square inch absolute (p), the 
vaporization temperature (t), the heat of the liquid (q, h or i'), 
the heat of vaporization (r or L), the specific volume (s or v), 

entropy of water (0 , 5 or n), entropy of vaporization ( — or — )> 

density pounds per cubic foot I- or -J, internal latent heat 

(p, i or I). In some tables will also be found a column for the 
total heat of vapor (H = q + r), external latent heat (APu) 
and entropy. 

Relation between Temperature, Pressure and Volume of 
Saturated Steam. The important relations of temperature, pres- 
sure and volume were first determined in a remarkable series of 
experiments conducted by a French engineer named Regnault, 
and it has been on the basis of his data, first published in 1847, 
that even our most modern steam tables were computed. Later 
experimenters have found, however, that these data were some- 
what in error, especially for values near the dry saturated 
condition. These errors resulted because it was difficult in the 
original apparatus to obtain steam entirely free from moisture. 

The pressure of saturated steam increases very rapidly as the 
temperature increases in the upper limits of the temperature scale. 
It is very interesting to examine a table of the properties of 
steam to observe how much more rapidly the pressure must be 
increased in the higher limits for a given range of temperature. 

It should be observed that in most tables the pressure is almost 
invariably given in terms of pounds per square inch, while in 
nearly ail our thermodynamic calculations the pressure must 
be used in pounds per square foot. 

The specific volume of saturated steam (v or s) is equal to 
the sum of the volumes of water (a) and of the increase in 
volume during vaporization («) , or 

V = a + U. (100) 



66 PROPERTIES OF VAPORS 

Heat in the Liquid (Water) (h or q). The essentials of the 
process of making steam have been described in a general way. 
The relation of this process to the amount of heat required will 
now be explained. If a pound of water which is initially at 
some temperature t is heated at a constant pressure P (pounds 
per square foot) to the boiling point corresponding to this pres- 
sure and then converted into steam, heat will first be absorbed 
in raising the . temperature of the water from to to t, and then 
in producing vaporization. During the first stage, while the 
temperature is rising, the amount of heat taken in is approxi- 
mately (t — t ) heat units, that is, British thermal units (B.t.u.), 
because the specific heat of water is approximately unity and 
practically constant. This number of B.t.u. multiplied by 778 
gives the equivalent number of foot-pounds of work. For the 
purpose of stating in steam tables the amount of heat required for 
this heating of water, the initial temperature t must be taken at 
some definite value; for convenience in numerical calculations 
and also because of long usage, the temperature 32 F. is invari- 
ably used as an arbitrary starting point for calculating the 
amount of heat " taken in." The symbol h (or sometimes i' or q) 
is used to designate the heat required to raise one pound of 
water from 32 F. to the temperature at which it is vaporized into 
steam. In other words, " the heat of the liquid" (h) is the 
amount of heat in B.t.u. required to raise one pound of water 
from 3 2 F. to the boiling point, or: 



h = \C dt. (101) 



C is the specific heat of water at constant pressure, U is the 
freezing point of water, / is the temperature to which the water 
is raised. 

As C is very nearly unity, the value of the heat absorbed by 
water in being raised to the steaming temperature (h) in B.t.u. 
can be expressed, approximately, by the formula 

h = t — 32 (in B.t.u.). (102) 

Values of h, taking into consideration the variation in the 



EXTERNAL WORK OF EVAPORATION 67 

specific heat of water, will be found in the usual steam tables 
(Table 3). 

During this first stage, before any steaming has occurred, prac- 
tically all the heat applied is used to increase the stock of internal 
energy. The amount of external work done by the expansion 
of water as a liquid is practically negligible. 

Latent Heat of Evaporation (L or r). In the second stage of the 
formation of steam as described, the water at the temperature t, 
corresponding to the pressure, is changed into steam at that 
temperature. Although there is no rise in temperature, very 
much heat is nevertheless required to produce this evaporation 
or vaporization. The heat taken in during this stage is the 
latent heat of steam. In other words, the latent heat of steam 
may be defined as the amount of heat which is taken in by a 
pound of water while it is changed into steam at constant pres- 
sure, the water having been previously heated up to the tem- 
perature at which steam forms. The symbol L (also sometimes 
r) is used to designate this latent heat of steam. Its value 
varies with the particular pressure at which steaming occurs 
being somewhat smaller in value at high pressures than at low. 

The latent heat (L or r) of saturated steam can be approxi- 
mately calculated by the formula: 

L = 97O.4 — O.655 Q — 212) — O.OOO45 (t — 2T2) 2 . (103) 

External Work of Evaporation (Apu). A part of the heat taken 
in during the " steaming " process is spent in doing external 
work. Only a small part of the heat taken in is represented 
by the external work done in making the steam in the boiler, 
and the remainder of the latent heat (L) goes to increase the 
internal energy of the steam. The amount of heat that goes 
into the performing of external work is equal to P (the pressure 
in pounds per square foot) times the change of volume (u) 
occurring when the water is changed into steam, or Pu. 

Example. At the usual temperatures of the working fluid in 
steam engines the volume of a pound of water a is about sV of a 
cubic foot. The external work done in making one . pound of 



68 PROPERTIES OF VAPORS 

steam having finally a volume of V (cubic feet) at a constant pres- 
sure P (pounds per square foot) may be written in foot-pounds: 

External work = Pu = P (V — -gV). C 1 ^) 

This last equation can be expressed in British thermal units 
(B.t.u.) by APu where A = yfg-. It is apparent also from this 
equation that the external work done in making steam is less at 
low pressure than at high,* because there is less resistance to over- 
come, or, in other words, P in equation (104) is less. The heat 
equivalent of the external work is, therefore, a smaller propor- 
tion of the heat added at low temperature than at high. 

Total Heat of Steam (H). The heat added during the process 
represented by the first and second stages in the formation of a 
pound of steam, as already described, is called the total heat of 
saturated steam or, for short, total heat of steam, and is repre- 
sented by the symbol H. Using the symbols already defined, 
we can write, per pound of steam, 

H = h + L (B.t.u.). (105) 

In other words, this total heat of steam is the amount of heat 
required to raise one pound of water from 32 F. to the tempera- 
ture of vaporization and to vaporize it into dry steam at that 
temperature under a constant pressure. 

Remembering that h for water is approximately equal to the 
temperature less 32 degrees corresponding to the pressure at 
which the steam is formed (t) the total heat of steam is approxi- 
mately, 

H = (t - 32) + L. (106) 

To illustrate the application of equation (106) calculate the 
total heat of steam being formed in a boiler at an absolute pres- 
sure of 1 1 5 pounds per square inch. 

* Although at the lower pressure the volume of a given weight of steam is 
greater than at a higher pressure, the change of pressure is relatively so much greater 
in the process of steam formation that the product of pressure and change of 
volume, P (V2 — Vi), which represents the external work done, is less for low 
pressure steam than for high. 



INTERNAL ENERGY OF EVAPORATION AND OF STEAM 69 

From the steam tables (Table 3) the temperature t of the 
steam at a pressure of 115 pounds per square inch absolute is 
338 F., the latent heat of vaporization is 880 B.t.u. per pound 
and the total heat of the steam is 1189 B.t.u. per pound. To 
check these values with equation (106), by substituting values of 
L and t, 

E = (338 - 3 2 ) + 880 = 1186 B.t.u. per pound. 

When steam is condensed under constant pressure, the process 
which was called the " second stage " is reversed and the amount 
of heat equal to the latent heat of evaporation (L) is given up 
during the change that occurs in the transformation from steam 
to water. 

Internal Energy of Evaporation and of Steam. It was ex- 
plained in a preceding paragraph that when steam is forming 
not all of the heat added goes into the internal or " intrinsic " 
energy of the steam, but that a part of it was spent in performing 
external work. If, then, the internal energy of evaporation is 
represented by the symbol I L , 

U-L-PZ-^. (107) 

This equation represents the increase in internal energy which 
takes place in the changing of a pound of water at the tem- 
perature t into steam at the same temperature. In all the 
formulas dealing with steam the state of water at 32 F. has been 
adopted as the arbitrary starting point from which the taking in 
of heat was calculated. This same arbitrary starting point is 
used also in expressing the amount of internal energy in the steam. 
This is the excess of the heat taken in over the external work done 
in the process. 

The total internal energy (L H ) of a pound of saturated steam at 
a pressure P in pounds per square foot is equal to the total heat 
(H) less the heat equivalent of the external work done; thus, 

i a = n-p {V ~/ t) - (108) 

778 



70 PROPERTIES OF VAPORS 

Such reference is made here to the internal energy of steam 
because it is very useful in calculating the heat taken in and 
rejected by steam during any stage of its expansion or com- 
pression. 

Heat taken in = increase of internal energy + external work 
done. 

When dealing with a compression instead of an expansion the 
last term above (external work) will be a negative value to indi- 
cate that work is done upon the steam instead of the steam doing 
work by expansion. 

The following problem shows the calculation of internal energy 
and external work: 

Example. A boiler is evaporating water into dry and satu- 
rated steam at a pressure of 300 pounds per square inch abso- 
lute. The feed Water enters the boiler at a temperature of 145 F. 

The internal energy of evaporation per pound of steam is 

I - l P ( V ~ ^ = Sn 3 °° X I44 ^' 551 ~ 7 ^ 

778 " ' 6 778 - 

= 811.3 — 85.3 = 726.0 B.t.u., 

or taken directly from saturated steam tables equals 726.8 B.t.u. 
The total internal energy supplied above 32 F. per pound of 
steam is 

I H = H - P ( V ~ ^ = 1204.1 - 85.3 = 1118.8 B.t.u., 

or taken directly from saturated steam tables equals 11 18.5 B.t.u. 
External work done above 32 F. per pound of steam as calcu- 
lated from steam tables is 

H — I H = 1204.1 — 1118.5 = 85.6 B.t.u. 

External work of evaporation per pound of steam as calcu- 
lated from the steam tables is 

L -I L = 811.3 - 726.8 = 84.5 B.t.u. 

The external work done in raising the temperature of a pound 
of water from 32 F. to the boiling point is 

85.6 - 84.5 = 1.1 B.t.u. 



WET STEAM 7 1 

The external work done in raising the temperature of a pound 
of water from 3 2° F. to 145 F. is 

300 X 144 (0.0163 — 0.01602) -p. 

£ ttfL ^ - = 0.01 B.t.u. 

778 

The external work done in forming the steam from water at 
145 F. is, then, 

84.5 + 1. 10 — 0.01 = 85.59 B.t.u. 

as compared with 85.3 B.t.u. as calculated from 

P (V ~ A) 
778 

It is to be noticed that the external work done during the addi- 
tion of heat when the liquid is raised to the vaporization tem- 
perature is small as compared with other values, and for most 
engineering work it is customary to assume that no external work 
is done during the addition of heat to the water. With this 
assumption 

I H = h + L- Piy ~ A) =H - (Z, - I L ). (109) 

778 

Steam Formed at Constant Volume. When saturated steam 
is made in a boiler at constant volume, as, for example, when the 
piping connections from the boiler to the engines are closed, then 
no external work is done, and all the heat taken in is converted 
into and appears as internal energy I H of the steam. This 
quantity is less than the total heat H of steam, representing its 
formation at constant pressure, by quantity P (V — ^V) -5- 778, 
where P represents the absolute pressure at which the steam is 
formed in pounds per square foot and V is the volume of a 
pound of steam at this pressure in cubic feet. 

Wet Steam. In all problems studied thus far, dealing with 
saturated steam, it has been assumed that the steaming process 
was complete and that the water had been completely converted 
into steam. In actual practice it is not at all unusual to have 
steam leaving boilers which is not perfectly and completely vapor- 



72 PROPERTIES OF VAPORS 

ized; in other words, the boilers are supplying to the engines a 
mixture of steam and water. This mixture we call wet steam. 
It is steam which carries actually in suspension minute particles 
of water, which remain thus in suspension almost indefinitely. 
The temperature of wet steam is always the same as that of dry 
saturated steam as given in the steam tables, so long as any 
steam remains uncondensed. 

The ratio of the weight of moisture or water in a pound of wet 
steam to a pound of completely saturated steam is called the 
degree of wetness; and when this ratio is expressed as a per 
cent, it is called the percentage of moisture or " per cent wet." 
If in a pound of wet steam there is 0.04 pound of water in suspen- 
sion or entrained, the steam is four per cent wet. 

Another term, called the quality of steam, which is usually 
expressed by the symbol x, is also frequently used to represent 
the condition of wet steam. Quality of steam may be denned 
as the proportion of the amount of dry or completely evaporated 
steam in a pound of wet steam. To illustrate with the example 
above, if there is 0.04 pound of water in a pound of wet steam; 
the quality in this case would be 1 — 0.04 or 0.96. In this case 
the quality of this steam is ninety-six one-hundredths. 

With this understanding of the nature of wet steam it is obvi- 
ous that latent heat of a pound of wet steam is xL. Similarly, 
the total heat of a pound of wet steam is h + xL, and the 
volume of a pound of wet steam is x (V — ^V) + ih — xV + ^ 
(1 — x), or approximately it is equal to xV, because the term A 
(1 — x) is negligibly small except in cases where the steam is so 
wet as to consist mostly of water. Similarly, the internal energy 
in a pound of wet steam is 

Superheated Steam. When the temperature of steam is higher 
than that corresponding to saturation as taken from the steam 
tables, and is, therefore, higher than the standard temperature 
corresponding to the pressure, the steam is said to be super- 



THE TOTAL HEAT OF SUPERHEATED STEAM 73 

heated. In this condition steam begins to depart and differ 
from its properties in the saturated condition, and when super- 
heated to a very high degree it begins to behave somewhat 
like a perfect gas. 

There are tables of the properties of superheated steam just 
as there are tables of saturated steam. (See Marks and Davis', 
Goodenough's or Peabody's Steam Tables.) When dealing with 
saturated steam there is always only one possible temperature and 
only one specific volume to be considered. With superheated 
steam, on the other hand, for a given pressure there may be any 
temperature above that of saturated steam, and corresponding to 
each temperature there will be, of course, definite values for 
specific volume and total heat. Like a perfect gas the specific 
volume or the cubic feet per pound increases with the increase in 
temperature. 

The total heat of superheated steam is obviously greater than 
the total heat of saturated steam.* Thus, since the total heat 
of dry saturated steam is, as before, h + L, the total heat of 
superheated steam with D degrees of superheat is 

Us = h + L + C p X D; (in) 

or if the temperature of the superheated steam is / sup . and / sa t. is 
the temperature of saturated steam corresponding to the pressure, 

H s = h + L + C p (4u P . — feat.). (112) 

H s (equations in and 112) is the total heat of superheated 
steam or is the amount of heat required to produce a pound of 
steam with the required degrees of superheat from water at 32 F. 

To obtain the amount of internal energy of a pound of steam 
(superheated) corresponding to this total heat as stated above, 
the external work expended in the steaming process must be 
subtracted; that is, 

* In practice steam is called dry saturated when it is exactly saturated and has 
no moisture. It is the condition known simply as saturated steam as regards the 
properties given in the ordinary steam tables. The dry saturated condition is the 
boundary between wet steam and superheated steam. 



74 PROPERTIES OF VAPORS 

I H = h + L - P (F Sat - ~ A) + Cp (/ sup , - feaO - jP(7sUP ~ Fsat - ) 

778 778 

= ft -- 1; — Lsp \tsup. fsat.j r 

7 

778 ^ 0/ 

The term ^ can be neglected in the equations above for prac- 
tically all engineering calculations as the maximum error from 
this is not likely to be more than one in one thousand or xV per 
cent.* The accuracy of our steam tables for values of latent and 
total heat is not established to any greater degree. Making 
this approximation, the equation above becomes 

Ih = h -f- L -f- Cp (/sup. — /sat.) — ( • ) = H s '-' 

\ 778 / 778 

If examination is made of the properties of superheated steam 
as given in Marks and Davis' or in Goodenough's Steam Tables 
for superheated steam at 165 pounds per square inch absolute 
pressure and 150 F. superheat, the following results are secured: 

Marks & Davis Goodenough 

Temperature (/) 516.0 516.1 

Volume (7, v) 3.43 34i 

Total Heat (H, or h\ or i) 1277.6 1280.8 

The values of h and L for saturated steam at this pressure are 
respectively 338.2 and 856.8. From the curves given, Fig. i7,{ 

* The error in the value of total internal energy due to neglecting the term -fa 
in the exercise worked out on page 70 is 0.85 B.t.u. per pound (one per cent of the 
external work), or an error of .076 per cent in the final result. 

f In their tables Marks and Davis represent the total heat of superheated 
steam by h. In this book the symbol Hs is used for greater clearness and to be 
consistent with the symbols used in the preceding formulas. They use also v for 
specific volume in place of V as above. 

J The curves for values of Cp, as given in Fig. 17, are for average and not for 
instantaneous values such as are given in Fig. 18, of Marks and Davis' Tables and 
Diagrams. Great caution must be observed in the use of curves of this kind. 
Those giving instantaneous values can only be used for the value of Cp in the form- 
ulas given for the total heat of superheated steam after the average value has been 
found by integrating the curve representing these values for a given pressure. 



DRYING OF STEAM BY THROTTLING OR WIRE-DRAWING 75 



the specific heat of superheated steam at constant pressure (C p ) 
is 0.552. From these data, 
E s = h + L + C v X 150 = 338.2 + 856.8 + 0.552 X 150 = 1277.7. 

This value agrees very nearly with that given in the tables as 
previously indicated. 



























































.66 














































































































'J4 














































































































M 














































































































.SO 














































































































■m 

* a J» 

"g oq .56 

S | 

| 1 .54 




















































































































































































































Lb 


s. , 


Vbs 




























































































- 227.2 


" 






A 0, .52 

3 ? 














































"170.4 
-142.2- 
-113.6 


4! 






















































a ^0 














































- 85.2 


» 












































I 


- 


-5C.8 


» 






48 


























| 


















1 ■ 

— 28.4 


,, 










































































[ 


■* - " 






.46 














































































































,.44 














































































































,.42 

























































Fig. 17. 



200 250 300 350 400 450 500 550 COO 650 700 750 

Temperature ° P 
- Mean Values of Cp Calculated by Integration from Knoblauch 
and Jakob's Data. 



V = \ 0.5962 T—p (1+0.0014^) (- 



■0.0833 



|> (114) 
IP 



The specific volume (V) is calculated from the empirical form- 
ula derived from experimental results and is expressed as follows: 

'150 ,300,000 

where p is in pounds per square inch, V is in cubic feet per pound 
and T = t + 460 is the absolute temperature on the Fahrenheit 
scale.* 

Drying of Steam by Throttling or Wire-drawing. When steam 
expands by passing through a very small opening, as, for example, 



* The value of 7 or ■=- of superheated steam used in ordinary engineering 
calculations is 1.3. 



7 6 



PROPERTIES OF VAPORS 



through a valve only partly open in a steam line, the pressure 
is considerably reduced. This effect is called throttling or wire- 
drawing. The result of expansion of this kind when the pressure 



O.90 




ioo° 



150 



-200 250 o 300 

temperature G. 



•350 



Fig. i8. — Values of the "True" Specific Heat of Superheated Steam. 

is reduced and no work is done is that if the steam is initially 
wet it will be drier and if it is initially dry or superheated the 
degree of superheat will be increased. The reason for this is 
that the total heat required to form a pound of dry saturated 
steam (H) is considerably less at low pressure than at high, but 
obviously the total quantity of heat in a pound of steam must be 



THROTTLING OR SUPERHEATING CALORIMETERS 77 

the same after wire- drawing as it was before, neglecting radiation. 
Now if steam is initially wet and the quality is represented by 
Xi, then the total heat in the steam is represented by fa plus XiLi, 
in which fa and Li represent the heat of the liquid and the latent 
heat of the steam at the initial pressure. If, also, the quality, 
heat of liquid and the latent heat of the steam after wire-drawing 
are represented respectively by X2, fa and Z2, then fa .+ X1L1 

= fa + X2L2, or 

X1L1 + fa - fa . N 

X2= - (115) 

This drying action of steam in passing through a small open- 
ing or an orifice is very well illustrated by steam discharging 
from a small leak in a high pressure boiler into the atmosphere. 
It will be observed that no moisture is visible in the steam a 
few inches from the leak but farther off it becomes condensed 
by loss of heat, becomes clouded and plainly visible. An im- 
portant application of the throttling principle is also to be found 
in the throttling calorimeter (page 80). 

Determination of the Moisture in Steam. Unless the steam 
used in the power plant is superheated it is said to be either dry 
or wet, depending on whether or not it contains water in sus- 
pension. The general types of steam calorimeters used to de- 
termine the amount of moisture in the steam may be classified 
under three heads: 

1 . Throttling or superheating calorimeters. 

2. Separating calorimeters. 

3. Condensing calorimeters. 

Throttling or Superheating Calorimeters. The type of steam 
calorimeter used most in engineering practice operates by passing 
a sample of the steam through a very small orifice, in which it 
is superheated by throttling. A very satisfactory calorimeter of 
this kind can be made of pipe fittings as illustrated in Fig. 19. 
It consists of an orifice O discharging into a chamber C, into 
which a thermometer T is inserted, and a mercury manometer 



78 PROPERTIES OF VAPORS 

is usually attached to the cock V 3 for observing the pressure in 
the calorimeter. 

It is most important that all parts of calorimeters of this 
type, as well as the connections leading to the main steam pipe, 
should be very thoroughly lagged by a covering of good insu- 
lating material. One of the best materials for this use is hair 
felt, and it is particularly well suited for covering the more or 
less temporary pipe fittings, valves and nipples through which 
steam is brought to the calorimeter. Throttling calorimeters 
have been found useless because the small pipes leading to the 
calorimeters were not properly lagged, so that there was too 
much radiation, producing, of course, condensation, and the 
calorimeter did not get a true sample. It is obvious that if the 
entering steam contains too much moisture the drying action 
due to the throttling in the orifice may not be sufficient to super- 
heat. It may be stated in general that unless there are about 
5° to io° F. of superheat in the calorimeter, or, in other words, 
unless the temperature on the low pressure side of the orifice is 
at least about 5 to io° F. higher than that corresponding to the 
pressure in the calorimeter, there may be some doubt as to the 
accuracy of results.* The working limits of throttling calo- 
rimeters vary with the initial pressure of the steam. For 35 
pounds per square inch absolute pressure the calorimeter ceases 
to superheat when the percentage of moisture exceeds about 
2 per cent; for 150 pounds absolute pressure when the moisture 
exceeds about 5 per cent; and for 250 pounds absolute pressure 
when it is in excess of about 7 per cent. For any given pressure 
the exact limit varies slightly, however, with the pressure in the 
calorimeter. 

* The same general statement may be made as regards determinations of 
superheat in engine and turbine tests. Experience has shown that tests made 
with from o to 10 degrees Fahrenheit superheat are not reliable, and that the 
steam consumption in many cases is not consistent when compared with results 
obtained with wet or more highly superheated steam. The errors mentioned 
when they occur are probably due to the fact that in steam, indicating less than 
10 degrees Fahrenheit superheat, water in the liquid state may be taken up in 
"slugs" and carried along without being entirely evaporated. 



THROTTLING OR SUPERHEATING CALORIMETERS 79 

In connection with a report on the standardizing of engine 
tests, the American Society of Mechanical Engineers * published 
instructions regarding the method to be used for obtaining a 
fair sample of the steam from the main pipes. It is recom- 
mended in this report that the calorimeter shall be con- 
nected with as short intermediate piping as possible with a 
so-called calorimeter nipple made of J-inch pipe and long enough 
to extend into the steam pipe to within | inch of the opposite 
wall. The end of this nipple is to be plugged so that the steam 
must enter through not less than twenty f-inch holes drilled 
around and along its length. None of these holes shall be less 
than J inch from the inner side of the steam pipe. The sample 
of steam should always be taken from a vertical pipe as near as 
possible to the engine, turbine or boiler being tested. A good 
example of a calorimeter nipple is illustrated in Fig. 20. 

The discharge valve V 2 should not be closed or adjusted 
without' first closing the gage cock V 3 . Unless this precaution is 
taken, the pressure may be suddenly increased in chamber C, so 
that if a manometer is used the mercury will be blown out of it; 
and if, on the other hand, a low-pressure steam gage is used it 
may be ruined by exposing it to a pressure much beyond its scale. 

Usually it is a safe rule to begin to take observations of tem- 
perature in calorimeters after the thermometer has indicated a 
maximum value and has again slightly receded from it. The 
quality or relative dryness of wet steam is easily calculated by 
the following method and symbols : 

pi = steam pressure in main, pounds per square inch 

absolute, 
p 2 = steam pressure in calorimeter, pounds per square 

inch absolute, 
t c = temperature in calorimeter, degrees Fahrenheit, 
L\ and hi = heat of vaporization and heat of liquid corre- 
sponding to pressure p h B.t.u., 
H 2 and t 2 = total heat (B.t.u.) and temperature (degrees 
Fahrenheit) corresponding to pressure p 2 , 
* Proceedings American Society of Mechanical Engineers, vol. XXI. 



8o 



PROPERTIES OF VAPORS 



Op 



specific heat of superheated steam. Assume 0.47 

for low pressures existing in calorimeters, 
initial quality of steam. 



Connection for 
Manometer 




Fig. 19. — Simple Throttling Steam Calorimeter. 

Total heat in a pound of wet steam flowing into orifice is 

XiU + h h 

and after expansion, assuming all the moisture is evaporated, 
the total heat of the same weight of steam is 

H 2 ~r Lp \pc — W). 

Then assuming no heat losses and putting for C p its value 0.47 : 

X1L1 + h = H 2 + 0.47 (t e — t 2 ) (116) 

H 2 + 0.47 (t c — t 2 ) — hi 



Xi = 



JU 



(117) 



BARRUS, THROTTLING CALORIMETER 8l 

The following example shows the calculations for finding the 
quality of steam from the observations taken with a throttling 
calorimeter : 

Example. Steam at a pressure of ioo pounds per square inch 
absolute passes through a throttling calorimeter. In the calo- 
rimeter the temperature of the steam becomes 243 ° F. and the 
pressure 15 pounds per square inch absolute. Find the quality. 

Solution. By taking values directly from tables of properties 
of superheated steam,* the total heat of the steam in the calorim- 
eter at 15 pounds per square inch absolute pressure and 243 ° F. 
is 1 164.8 B.t.u. per pound or can be calculated as follows: 

E + C v {t c - t 2 ) = 1150.7 + 0.47 ( 2 43 - 2I 3) 
= 1 164.8 B.t.u. per pound. 

(Note. t c = temperature in calorimeter and t 2 = temperature 
corresponding to calorimeter pressure.) 

The total heat of the steam before entering the calorimeter 
is h + xL. At 100 pounds per square inch absolute pressure, 
this is 298.3 + 888.0 x in B.t.u. per pound. Since the heat in 
the steam per pound in the calorimeter is obviously the same as 
before it entered the instrument, we can equate as follows: 

298.3 + 888.0 x = 1164.8 
x = 0.976 
or the steam is 2.4 per cent wet. 

Barrus Throttling Calorimeter. This is an important varia- 
tion from the type of throttling calorimeter shown in Fig. 19 and 
has been quite widely introduced by Mr. George H. Barrus. In 
this apparatus the temperature of the steam admitted to the 
calorimeter is observed instead of the pressure and a very free 
exhaust is provided so that the pressure in the calorimeter is 
atmospheric. This arrangement simplifies very much the obser- 
vations to be taken, as the quality of the steam X\ can be calcu- 
lated by equation (117) by observing only the two temperatures 

* Marks and Davis' Steam Tables and Diagrams (1st ed.), page 24 or 
Goodenough's Properties of Steam and Ammonia (2d ed.), page 47. 



82 



PROPERTIES OF VAPORS 




Fig. 20. 



Barms' Throttling Steam 
Calorimeter. 



/1 and t c , taken respectively on the high and low pressure sides of 
the orifice in the calorimeter. This calorimeter is illustrated in 
Fig. 20. The two thermometers required are shown in the 
figure. Arrows indicate the path of the steam.* 

The orifice in such calorimeters is usually made about ft- inch 
in diameter, and for this size of orifice the weight of steam | 

discharged per hour at 175 pounds 
per square inch absolute pressure 
is about 60 pounds. It is impor- 
tant that the orifice should always 
be kept clean, because if it be- 
comes obstructed there will be a 
reduced quantity of steam passing 
through the instrument, making 
the error due to radiation rela- 
tively more important. 

In order to free the orifice from 
dirt or other obstructions the connecting pipe to be used for 
attaching the calorimeter to the main steam pipe should be 
blown out thoroughly with steam before the calorimeter is put 
in place. The connecting pipe and valve should be covered 
with hair felting not less than f inch thick. It is desirable also 
that there should be no leak at any point about the apparatus, 
as in the stuffing-box of the supply valve, the pipe joints or the 
union. 

With the help of a diagramf giving the quality of steam directly 
the Barrus calorimeter is particularly well suited for use in 
power plants, where the quality of the steam is entered regu- 
larly on the log sheets. The percentage of moisture is obtained 
immediately from two observations without any calculations. 

Separating Calorimeters. It was explained that throttling 
calorimeters cannot be used for the determination of the quality 
of steam when for comparatively low pressures the moisture 

* Transactions of American Society of Mechanical Engineers, vol. XI, page 790. 
f In boiler tests corrections should be made for the steam discharged from the 
steam calorimeters. 

J See page 113, and also Moyer's Power Plant Testing, (2d ed.), pages 58-60. 



SEPARATING CALORIMETERS 



83 



-z = Steam 

Water 



is in excess of 2 per cent, and when for boiler pressures commonly 
used it exceeds 5 per cent. For higher percentages of moisture 
than these low limits separating calorimeters are most generally 
used. In these instruments the water is removed from the 
sample of steam by mechanical separation just as it is done 
in the ordinary steam separator installed in the steam mains 
of a power plant. There is pro- 
vided, of 'course, a device for de- 
termining, while the calorimeter is 
in operation, usually by means of 
a calibrated gage glass, the amount 
of moisture collected. This me- 
chanical separation depends for its 
action on changing very abruptly 
the direction of flow and reducing 
the velocity of the wet steam. 
Then, since the moisture (water) 
is nearly 300 times as heavy as 
steam at the usual pressures de- 
livered to the engine, the moisture 
will be deposited because of its 
greater inertia. 

Fig. 21 illustrates a form of sep- 
arating calorimeter having a steam 
jacketing spaqe which receives live 

steam at the same temperature as the sample. Steam is supplied 
through a pipe A, discharging into a cup B. Here the direction 
of the flow is changed through nearly 180 degrees, causing the 
moisture to be thrown outward through the meshes in the cup 
into the vessel V. The dry steam passes upward through the 
spaces between the webs W, into the top of the outside jacketing 
chamber J, and is finally discharged from the bottom of this 
steam jacket through the nozzle N. This nozzle is consider- 
ably smaller than any other section through which the steam 
flows, so that there is no appreciable difference between the pres- 
sures in the calorimeter proper and the jacket. The scale 




Fig. 21. — Separating Calorimeter. 



84 PROPERTIES OF VAPORS 

opposite the gage glass G is graduated to show, in hundredths 
of a pound, at the temperature corresponding to steam at 
ordinary working pressures, the variation of the level of the 
water accumulating. A steam pressure gage P indicates the 
pressure in the jacket J, and since the flow of steam through 
the nozzle N is roughly proportional to the pressure, another 
scale in addition to the one reading pressures is provided 
at the outer edge of the dial. A petcock C is used for draining 
the water from the instrument, and by weighing the water col- 
lected corresponding to a given difference in the level in the gage 
G the graduated scale can be readily calibrated. Too much 
reliance should not be placed on the readings for the flow of steam 
as indicated by the gage P unless it is frequently calibrated. 
Usually it is very little trouble to connect a tube to the nozzle N 
and condense the steam discharged in a large pail nearly filled 
with water. When a test for quality is to be made by this method 
the pail nearly filled with cold water is carefully weighed, and then 
at the moment when the level of the water in the water gage G 
has been observed the tube attached to the nozzle N is immedi- 
ately placed under the surface of the water in the pail. The 
test should be stopped before the water gets so hot that some 
weight is lost by " steaming." The gage P is generally cali- 
brated to read pounds of steam flowing in ten minutes. For the 
best accuracy it is desirable to use a pail with a tightly fitting 
cover into which a hole just the size of the tube has been cut. 

If W is the weight of dry steam flowing through the orifice N 
and w is the weight of moisture separated, the quality of the 
steam is 

W + w 

Condensing or Barrel Calorimeter. For steam having a large 
percentage of moisture (over 5 per cent) the condensing or 
barrel calorimeter will give fairly good results if properly used. 
In its simplest form it consists of a barrel placed on a platform 
scale and containing a known weight of cold water. The steam 
is introduced by a pipe reaching nearly to the bottom of the 



EQUIVALENT EVAPORATION AND FACTOR OF EVAPORATION 85 

barrel. The condensation of the steam raises the temperature 
of the water, the loss of heat by the wet steam being equal to 
the gain of heat by the cold water. It will, therefore, be neces- 
sary to observe the initial and final weights, the initial and 
final temperatures of the water in the barrel and the temperature 
of the steam. 

Let W = original weight of cold water, pounds. 
w = weight of wet steam introduced, pounds. 
tx — temperature of cold water, degrees Fahrenheit. 
t 2 = temperature of water after introducing steam. 
t s = temperature of steam. 
L = latent heat of steam at temperature t s . 
x = quality of steam. 

Then, 

Heat lost by wet steam = heat gained by water. 

wxL + w (t s — h) = W (t 2 — h) 

X = wl (lI9) 

The accuracy of the results depends obviously upon the accu- 
racy of the observation, upon thorough stirring of the water so 
that a uniform temperature is obtained and upon the length of 
time required. The time should be just long enough to obtain 
accurate differences in weights and temperatures; otherwise, 
losses by radiation will make the results much too low. 

Equivalent Evaporation and Factor of Evaporation. For the 
comparison of the total amounts of heat used for generating 
steam (saturated or superheated) under unlike conditions it. is 
necessary to take into account the temperature t Q at which the 
water is put into the boiler as well as also the pressure P at 
which the steam is formed.* These data are of much impor- 
tance in comparing the results of steam boiler tests. The basis 

* As the pressure P increases the total heat of the steam also increases; but 
as the initial temperature of the water ("feed temperature") increases the value 
of the heat of the liquid decreases. 



86 PROPERTIES OF VAPORS 

of this comparison is the condition of water initially at the boil- 
ing point for "atmospheric" pressure or at 14.7 pounds per 
square inch; that is, at 212 F. and with steaming taking place 
at the same temperature. For this standard condition, then, 
h = o and H = L = 970.4 B.t.u. per pound. Evaporation 
under these conditions is described as, 

" from (a feed-water temperature of) and at (a pressure corre- 
sponding to the temperature of) 212 F." 

To illustrate the application of a comparison with this stand- 
ard condition, let it be required to compare it with the amount 
of heat required to generate steam at a pressure of 200 pounds 
per square inch absolute with the temperature of the water sup- 
plied (feed water) at 190 F. 

For P = 200 pounds per square inch absolute the heat of the 
liquid h = 354.9 B.t.u. per pound and the heat of evaporation 
(L) is 843.2 B.t.u. per pound. For t = 190 F. the heat of the 
liquid (ho) is 157.9 B.t.u. per pound. The total heat actually 
required in generating steam at these conditions is, therefore, 

843.2 + (354.9 — I57-9) = 1040-2 B.t.u. per pound. 

The ratio of the total heat actually used for evaporation to 
that necessary for the condition defined by " from and at 212 F." 
is called the factor of evaporation. In this case it is the value 

1040.2 t- 970.4 = 1.07. 

Calling F the factor of evaporation, h and L respectively the 
heats of the liquid and of evaporation corresponding to the steam 
pressure and h Q the heat of the liquid corresponding to the tem- 
perature of feed water, then 

^ = L + (h-ho) . (i2q) 

970.4 

The actual evaporation of a boiler (expressed usually in pounds 
of steam per hour) multiplied by the factor of evaporation is 
called the equivalent evaporation. 



PROBLEMS 87 

Vapors as Refrigerating Media. Ammonia is generally used 
as the refrigerating medium in connection with mechanical 
refrigeration. Carbon dioxide and sulphur dioxide are also used 
to a limited extent as refrigerating media. 

The value of a substance as a refrigerating medium depends 
upon its latent heat, its vaporization temperature, its cost, and 
upon its chemical properties. The greater the latent heat of a 
refrigerating medium the more heat will it be capable of abstract- 
ing by evaporation. Upon the vaporization temperature of the 
refrigerant at different pressures depends the degree of cold it 
can produce as well as its practicability for use in hot climates 
where low temperature cooling water cannot be secured. 

Goodenough's Tables of Properties of Steam and Ammonia give 
the various physical properties of saturated, superheated and 
liquid ammonia. Peabody's Tables give the properties of 
ammonia, sulphur dioxide, carbon dioxide and of other vapors 
used as refrigerating media. 

Table 4 is an abridged table based upon Goodenough's values 
for saturated ammonia. Tables 5 and 6, based upon Peabody's 
Tables, show some of the properties of sulphur dioxide (S0 2 ) and 
of carbon dioxide (C0 2 ). 

PROBLEMS 

1. Dry and saturated steam has a pressure of 100 lbs. per sq. in. absolute. 
Calculate the temperature of the steam, the volume per pound, the heat 
of the liquid, the latent heat, and the total heat above 32 F. per pound of 
this steam. 

2. Dry and saturated steam has a temperature of 300 F. What is 
its pressure, heat of the liquid, latent heat, and total heat? 

3. A closed tank contains 9 cu. ft. of dry and saturated steam at a 
pressure of 150 lbs. per sq. in. absolute. 

(a) What is its temperature? 

(b) How many pounds of steam does the tank contain? 

4. A boiler generates dry and saturated steam under a pressure of 200 
lbs. per sq. in. absolute. The feed water enters the boiler at 6o° F. 

(a) What is the temperature of the steam? 

(b) How many British thermal units are required to generate 

1 lb. of this steam if this feed water is admitted at 32 F.? 



88 PROPERTIES OF VAPORS 

(c) How many British thermal units are required to generate 
i lb. of this steam from feed water at 60 ° F. into the 
steam at the pressure stated at the beginning of this prob- 
lem? 

5. One pound of dry and saturated steam is at a pressure of 250 lbs. 
per sq. in. absolute. 

(a) What is its internal energy of evaporation? 

(b) What is its total internal energy above 32 F.? 

(c) How much external work was done during its formation from 

32° F.? 

(d) How much external work was done during the evaporation? 

(e) How much external work was done during the change in 

temperature of the water from 32 degrees to the boiling 
point corresponding to the pressure? 

6. Dry and saturated steam is generated in a boiler and has a tempera- 
ture of 400 F. The feed water enters the boiler at 200 F. 

(a) What pressure is carried in the boiler? 

(b) What is the total heat supplied to generate 1 lb. of this steam? 

(c) How much external work was done during its formation? 

(d) How much heat was used in increasing the internal energy? 
Check this by (hi — hi + II) noting that this assumes no 

external work done in the heating of the liquid. 

7. One pound of steam at a pressure of 100 lbs. per sq. in. absolute has 
a quality of 90 per cent dry. What is its temperature? 

8. How many British thermal units would be required to raise the 
pound of steam in the above problem from 32 F. to the boiling point corre- 
sponding to the pressure stated? 

9. What would be the volume of a pound of steam for the conditions 
stated in problem 7? 

10. How many heat units (latent heat) are required to evaporate the 
steam in problem 7? 

11. What is the amount of the total heat (above 32 F.) of the steam in 
problem 7? 

12. What would be the external work of evaporation of the steam in 
problem 7? 

13. How much external work (above 32 F.) is done in making steam as 
in problem 7? 

14. What is the internal energy of evaporation of the steam in problem 7? 

15. What is the total internal energy of the steam in problem 7? 

16. A tank contains 9 cu. ft. of steam at 100 lbs. per sq. in. absolute 
pressure which has a quality of 0.95. How many pounds of steam does the 
tank contain? 



PROBLEMS 89 

17. Two pounds of steam have a volume of 8 cu. ft. at a pressure of 100 
lbs. per sq. in. absolute. What is the quality? 

Calculate its total heat above 32 F. 

18. One pound of steam having a quality of 0.95 has a temperature of 
3 2 5 F. What is the pressure? 

19. One pound of steam at a pressure of 225 lbs. per sq. in. absolute has 
a temperature of 441. 9 F. 

(a) Is it superheated or saturated? 

(b) What is the total heat required to generate such steam from 

water at 32 F.? 

(c) What is its volume? 

(d) How much external work was done (above 32 F.) in generat- 

ing it? 

(e) How much internal energy above 32 F. does it contain? 

20. One pound of steam at a pressure of 300 lbs. per sq. in. absolute has 
a volume of 1.80 cu. ft. 

(a) Is it saturated or superheated? 

(b) What is its temperature? 

(c) How much is its total heat above 32 F.? 

(d) What is its total internal energy? 

21. Steam in a steam pipe has pressure of 110.3 lbs. per sq. in. by the 
gage. A thermometer in the steam registers 385 F. Atmospheric pres- 
sure is 14 lbs. per sq. in. absolute. Is the steam superheated, and if 
superheated how many degrees? 

22. Steam at a pressure of 200 lbs. per sq. in. absolute passes through 
a throttling calorimeter.. After expansion into the calorimeter the tem- 
perature of this steam is 250 F. and the pressure 15 lbs. per sq. in. abso- 
lute. What is the quality? 

23. Steam at a temperature of 325 F. passes through a throttling 
calorimeter. In the calorimeter the steam has a pressure of 16 lbs. per 
sq. in. absolute and a temperature of 236. 3 F. What is the quality? 

24. Steam at 150 lbs. per sq. in. absolute pressure passes through a 
throttling calorimeter. Assuming that the lowest condition in the calorim- 
eter for measuring the quality is io° F. superheat and the pressure in the 
calorimeter is 15 lbs. per sq. in. absolute, what is the largest percentage of 
wetness the calorimeter is capable of measuring under the above conditions? 

25. Prepare a chart by means of which can be determined the largest 
percentages of wetness a throttling calorimeter will measure at all pressures 
from 50 to 300 lbs. per sq. in. absolute. 

26. In a ten-minute test of a separating calorimeter the quantity of dry 
steam passing through the orifice was 9 pounds. The quantity of water 
separated was 1 pound. What was the quality? 



90 PROPERTIES OF VAPORS 

27. A barrel contained 400 lbs. of water at a temperature of 50 F. Into 
this water steam at a pressure of 125 lbs. per sq. in. absolute was admitted 
until the temperature of the water and condensed steam in the barrel 
reached a temperature of ioo° F. The weight of the water in the barrel 
was then 418.5 lbs. What was the quality? 

28. Calculate the factor of evaporation and the equivalent evaporation 
per pound of coal for a boiler under the following conditions: Steam pres- 
sure, 190 lbs. per sq. in. gage; feed water temperature, 203 F.; steam appar- 
ently evaporated per pound of coal i\ pounds; steam 3 per cent wet. 

29. Plot on one chart the pressure-temperature relations of ammonia, 
SO2, and CO2. 

30. Plot on one chart the pressure-latent heat relations of ammonia, 
SO2, and CO2. 

31. What conclusions do you reach from the graphic results in problems 
29 and 30? 

32. If cooling water for condensing ammonia cannot be secured at a tem- 
perature less than 90 F., what will be the pressure of the ammonia at the 
inlet to the condenser of the refrigerating system? 



CHAPTER VI 



t 

I 

■3 

w 








SQ 



/ 



dQ 



T .i 

Entropy- 4> 

- Analysis of Entropy 
Diagram. • 



ENTROPY 

Pressure-volume diagrams are useful for determining the 
work (in foot-pounds), done during any process or a cycle, but 
they are of very limited use in analyz- 
ing the heat changes involved. It 
has, therefore, been found desirable to 
make use of a diagram which shows 
directly by an area the number of 
heat units (instead of foot-pounds) in- 
volved during any process. In order 
that an area shall represent this value 
the coordinates must be such that their 
product will give heat units. If the 
ordinates are in absolute temperature, Fig. 22. 
the abscissas must be heat units per 

degree of absolute temperature, that is, -^> for then T X^ = Q, 

the amount of heat added during the process. 

Suppose that the heat Q, which is required to produce a process, 
such as an expansion or a compression, is divided up into a 
number of small increments dQ (Fig. 22) and that each small 
increment of heat is divided by the average absolute temperature 
at which the heat change occurs. There will then be a series 

of expressions -^ which, when integrated, will give the total 

change in the abscissas. This quantity I -^ when multiplied by 

the average absolute temperature between C and D will give the 
total amount of heat involved during the process. 
Mathematically expressed, the change in the abscissas is 

dQ 
T 
- 



v T 



or <j> = 



f 



(121) 



92 ENTROPY 

The heat change involved is 

dQ = Td<j> or Q = Ct d<t>. (122) 

The quantity <j> in the equations is known as the increase in en- 
tropy of the substance, and may be denned as a quantity which, 
when multiplied by the average absolute temperature occurring 
during a process, will give the number of heat units (in B.t.u.) 
added or abstracted as heat during the process. The " increase 
in entropy" is employed rather than entropy itself, because 
only the differences in entropy are important. 

This definition of entropy states that in a temperature-entropy 
diagram such as Fig. 22, where the ordinates are absolute tem- 
peratures, and the abscissas are entropies as calculated above 
some standard temperature, the area under any line CD gives the 
number of heat units added to the substance in passing from a 
temperature T\ and entropy <£i = Oe to a temperature T 2 and 
entropy <£ 2 = Of (or the number of heat units abstracted in pass- 
ing from T 2 to 7\) . 

Entropy Changes During Constant Pressure Expansions of 
Gases. The heat supplied or abstracted during a constant 
pressure expansion or compression of a gas may be stated: 

dQ = wCpdt, (123) 

or (^ = wC p (r 2 — Ti). (124) 

From equation (121) the change in entropy is 

Assuming the specific heat C v constant, the change in entropy 

becomes : 

r T *dt 
= wCp J - (126) 

= WC P (loge T 2 — l0g e Ti) 

= wC P \og e7 ^- (127) 

I 1 



ENTROPY CHANGES 93 

Entropy Changes of Gases at Constant Volume. The heat 
supplied or abstracted during a constant volume change of a gas 
may be stated: 

dQ = wCvdt, (128) 

or Q = wC v (T 2 — Ti). (129) 

Assuming, as before, the specific heat under constant volume 
(C v ) constant and substituting in equation (121), the change in 
entropy is 

= wC v J - (130) 

= WCv (l0g e T 2 - lOge Ti) (131) 

= wC v log e7 ^' (132) 

1 1 

Entropy Changes During Isothermal Processes of a Gas. 

During an isothermal change the temperature remains constant. 
The heat supplied or abstracted is consequently equal to the 
heat equivalent of the work done, which in terms of weight and 
temperature is 

wRTloge-^ ? - (133) 

Vl 



Since the temperature remains constant, the change in entropy 



is 



= 



«* ri °*£; (134) 



= WR \0g e ^. (135) 

V 1 

Entropy Changes During Reversible Adiabatic Processes of 
Gases. The heat supplied to or abstracted from an adiabatic 
expansion or compression is, by definition, zero. The entropy 
change in such a process is 

*=r?=°- (i36) 

Hence during an adiabatic process no change of entropy occurs. 



"7 
d - 

s 


T X A B 








H - 


T 2 


D 


c 






(V 








■»-> 








3 








f-H 








o 








OQ 








.Q 












e 


/ 



94 ENTROPY 

It is possible for adiabatic processes to occur which are not 
reversible. These are processes in which, not only no heat is 
added or abstracted, but a portion or all the work of the process 
may reappear in the working medium as heat. In such a process 
the entropy does not remain constant. Reversible adiabatic 
processes are sometimes called isoentropic (equal entropy) to 
distinguish them from the irreversible adiabatic processes.^ 

Entropy Changes During a Carnot Cycle. A pound of the 
working substance is first expanded isothermally. On a T-cj> 

(temperature-entropy) diagram (Fig. 
23), this process would be represented 
by line AB, where the temperature re- 
mains constant at T\, and where the 
entropy increases from Oe to Of, be- 
cause of the addition of heat that is re- 
Entropy-0 quired to keep the temperature constant. 

Fig. 23. — Entropy Diagram The next process is adiabatic expan- 

of Carnot Cycle. • r t> 4. 't tt ^ • *i_i 

3 sion from ii to 1 2 - Heat is neither 

added to nor abstracted from the substance during this expan- 
sion. Hence the entropy remains constant, as indicated by 
BC. 

The substance is now isothermally compressed along CD, the 
temperature remaining constant at T 2 and the entropy decreasing 
because of the abstraction of heat. 

The last process of the cycle is the adiabatic compression from 
D to A, no heat being added or abstracted, and the entropy, 
therefore, remaining constant. 

The heat supplied during the cycle is represented by the area, 
ABfe and equals 

($f - 4>e) Tl (137) 

The heat abstracted during the cycle is represented by the 
area, CDef, and equals 

(0/ - e ) T 2 . (138) 

Since the work of the cycle, in terms of heat, equals the heat 
added minus the heat rejected, 



TEMPERATURE-ENTROPY DIAGRAMS FOR STEAM 95 

Work = Area ABfe — area CDef 
= Area ABCD. 

In terms of entropy, 

Work = (</>/ — <f> e ) Ti — (</>/ — 4> e ) T 2 

= (7\ - T 2 ) (0/ - e ). (139) 

Efficiency of cycle is, 

„ Area ABCD 
E = 



Area ABfe 
(7\ - - T 2 ) (0/ - 4>c) = T, - T 2 
T\ (0/ — 0e) 2\ 



(140) 



From the foregoing discussion two important conclusions may 
be drawn in regard to the use of the T-<$> diagram : 

1. If any heat process be represented by a curve on a T-<f> 
diagram, the heat involved during the process is equal to the 
area under the curve, that is, between the curve and the axis of 
absolute temperature. 

2. If a cycle of heat processes be represented on a T-0 
diagram by a closed figure, the net work done is equal to the 
enclosed area, that is, the enclosed area measures the amount 
of heat that was converted into work. 

T-0 diagrams are useful for analyzing heat processes, and find 
particularly useful application in steam engineering, as indicated 
by the following: 

1. Graphical analysis of heat transfers in a steam engine 
cylinder. 

2. Determination of quality of steam during adiabatic ex- 
pansion. 

3. Calculations of steam engine cycle efficiencies. 

4. Steam turbine calculations. 

Temperature-Entropy Diagrams for Steam. A temperature- 
entropy diagram for steam is shown in Fig. 24. The various 
shaded areas represent the heats added to water at 32 F. to 
completely vaporize it at the pressure Pi. The area ABCD is 
the heat added to the water to bring it to the temperature of 
vaporization, or represents the heat of the liquid (h) for the pres- 



9 6 



ENTROPY 



sure P\. Further heating produces evaporation at the constant 
temperature T\ corresponding to the pressure Pi, and is repre- 
sented by the area under the line CE. When vaporization is 
complete, the latent heat, or the heat of vaporization (Z), is the 
area DCEF. If, after all the water is vaporized, more heat is 
added, the steam becomes superheated, and the additional heat 
required would be represented by an area to the right of E. 



Dry Steam or 
Saturation Line 




1.5 



2.0 



0.5 1.0 

Entropy (0) 

Fig. 24. — Temperature-entropy Diagram of Steam. 

Calculation of Entropy for Steam. In order to lay off the 
increase in entropy as abscissas in the heat diagram for steam, 
it is necessary to determine various values. For convenience 
3 2 F. has been adopted as the arbitrary starting point for cal- 
culating increase of entropy, as well as for the other thermal 
properties of steam. Referring to Fig. 25 the entropy of water 
is seen to be o at 32 F. In order to raise the temperature from 
To to Ti, h heat units are required, and by the equation (121), 
the entropy of one pound of the liquid 6 will be 

C Tl dh ( , 

= / -57> 04i) 



To 



T 



or, assuming the specific heat of water to be unity we can write 



T 

= l0ge T X - l0g e To = l0g e —-- 

1 



(142) 



CALCULATION OF ENTROPY FOR STEAM 



97 



In order to evaporate the water into steam at the boiling point 
Tij the latent heat, L, must be added at constant temperature 
7\. The increase of entropy during the "steaming" process is 

represented by — and is equal to, in this case, 



Ora. 

Eahr. 

328+788 



(143) 





k ft 

Entropy- 
logr e T 2 — H 

log e Tx- 
Fig. 25. — Diagram for Calculation of Entropy of Steam. 

Further heating would produce superheated steam and the 
change in entropy would be 



Entropy of superheat = / 

= C p log 



rsu p- Cpdt 

Ts&t. * ' 



T 

J- sup. 



sat. 



(144) 
(145) 



where C P = specific heat of the superheated steam at constant 
pressure. 

T S uv. = absolute temperature of the superheated steam. 

Ts*t. = absolute temperature of the saturated steam. 



98 * ENTROPY 

Total Entropy of steam, which is the sum of entropies, corre- 
sponding to the various increments making up the total heat of 
the steam, depends upon the quality of the steam. 

For dry saturated steam the total entropy above 32 F. and 
temperature 7\ is equal to the sum of the entropy of the liquid 
and the entropy of evaporation for dry saturated steam. This 
may be written 

0sat. = #1 + — (l46) 

J- 1 

= log e -^ + -^ (147) 

i J- 1 

Similarly, for wet steam whose quality is x, the total entropy is 

0wet = 01 + -=- (l48) 

J- 1 

, 7\ xLi 

J- i 1 

For superheated steam whose final temperature is T sup . and 
the temperature corresponding to saturation is 7\, the total 
entropy is, by equation (144), 

*_-* + £ +■/•*■*£* • ■ ( I49 ) 

1 1 «^r sa t. J- 

= loge ^ + -£ + C p \0ge -^ ' (i 50) 

1 2 J- I JL sat. 

Values of these entropies will be found in steam tables. 

Referring to Fig. 25, the line Tic represents the increase of 
entropy due to the latent heat added during the steaming process. 
If this steaming process had stopped at some point such that 
the steam was wet, having a quality x, this condition of the steam 
could be denoted by the point s, where 

This relation is obvious, for a distance along Tic represents 
the entropy of steam, which is proportional to the latent heat 
added, which in turn is proportional to the amount of dry steam 



THE MOLLIER CHART 99 

rrt 

formed from one pound of water. In like manner, — ^— is the 

T 2 d 

quality of steam at the point m that has been formed along T 2 d 

at the temperature T 2 . 

It is thus apparent that any point on a T-cj> diagram will give 
full information in regard to the steam. The proportional dis- 
tances on a line drawn through the given point between the 
water and the dry steam lines and parallel to the X-axis give 
the quality of the steam as shown above. The ordinate of the 
point gives the temperature and corresponding pressure, while 
its position relative to other lines, such as constant volume and 
constant total heat lines which can be drawn on the same dia- 
gram, will give further important data. 

The Mollier Chart. While the temperature-entropy diagram 
is extremely useful in analyzing various heat processes, its use. 
in representing the various conditions of steam is not so conven- 
ient as that of the Mollier chart. This chart is illustrated in 
Plate 1 of the Appendix. The coordinates consist of the total 
heat of the steam above 32 F. and entropy. The saturation 
curve marks the boundary between the superheated and satu- 
rated regions. In the wet region lines of constant volume are 
drawn and in the superheated regions the lines of equal super- 
heats appear. In both the superheated and saturated regions, 
lines of constant pressure are drawn and the absolute pressures 
of the various curves are labeled. 

By means of the Mollier chart (Appendix) the curve for 
adiabatic expansion can be very readily drawn on a pressure- 
volume diagram when the initial quality of steam at cut-off 
is known. Assume that one pound of wet steam at a pressure 
of 150 pounds per square inch absolute and quality 0.965 dry 
is admitted to the engine cylinder per stroke, and that there 
is previously in the clearance space 0.2 cubic foot of steam 
(see Fig. 26), at exactly the same condition. The volume of a 
pound of dry saturated steam at this initial pressure of 1 50 
pounds is, from the steam-tables, 3.012 cubic feet. At 0.965 
quality it will be 1 (0.965 X 3.012) = 2.905 cubic feet. On the 



IOO 



ENTROPY 



scale of abscissas this amount added to the 0.20 cubic foot in 
the clearance gives 3.105 cubic feet, the volume to be plotted at 
cut-off. Other points in the adiabatic expansion curve can be 
readily plotted after determining the quality by the method given 
on page no. 



^7 Clearance Steam 
Admission 



t Cut-off 




0.20 3.15 

Volume of Steam in Cylinder (Cu. ft.) 

Fig. 26. — Illustrative Indicator Diagram of Engine using Steam with Expansion. 

Temperature-entropy Diagram for the Steam Power Plant. 
Fig. 27 illustrates the heat process going on when feed water 
is received in the boilers of a power plant at ioo° F., is heated 
and converted into steam at a temperature of 400 F., and then 
loses heat in doing work. When the feed water first enters 
the boiler its temperature must be raised from ioo° to 400 F. 
before any " steaming" begins. The heat added to the liquid 
is the area MNCD. This area represents the difference between 
the heats of the liquid (374 — 68) or about 306 B.t.u. The 
horizontal or entropy scale shows that the difference in entropy 
between water at ioo° and 400 F. is about 0.436.* 

The curve NC is constructed by plotting from the steam 
tables the values of the entropy of the liquid for a number of 
different temperatures between ioo° and 400 F. 

If water at 400 F. is converted into steam at that temperature, 

* As actually determined from Marks and Davis' Steam Tables (pages 9 and 
15), the difference in entropy is 0.5663 — 0.1295 or 0.4368. Practically it is im- 
possible to construct the scales in the figure very accurately. 



TEMPERATURE-ENTROPY DIAGRAM 



IOI 



the curve representing the change is necessarily a constant tem- 
perature line and therefore a horizontal, CE. Provided the 
evaporation has been complete, the heat added in the " steaming " 
process is the latent heat or the heat of evaporation of steam 
(L) at 400 F., which is approximately 827 B.t.u. 



400 • 



•5100 

§ 32! 
u 

1 

C3 




-460 



M 



Supei heated 
Steam 




0.130 



.566 1.528 

Entropy — <P 



Fig. 27. — Temperature-entropy Diagram for the Steam Power Plant. 

The change in entropy during evaporation is, then, the heat 
units added (827) divided by the absolute temperature at which 
the change occurs (400 -f- 460 = 86o° F. absolute) or 

L = 827 = 
T 860 



0.962. 



The total entropy of steam completely evaporated at 400 F. 
is, therefore, 0.566 + 0.962, or 1.528.* To represent this final 
condition of the steam, the point E is plotted when entropy 
measured on the horizontal scale is 1.528 as shown in the figure. 
The point E is shown located on another curve RS, which is 
determined by plotting a series of points calculated the same as 
E, but for different pressures. The area MNCEF represents the 
total heat added to a pound of feed water at 100 F., to produce 

* Entropy, like the total heat (H) and the heat of the liquid (h), is measured 
above the condition of freezing water (32 F.). 



102 ENTROPY 

steam at 400 F. Area OBCEF represents the total heat of 
steam (H in the steam tables) above 32 F. required to form one 
pound of steam at 400 F. 

If the steam generated in the boiler is now allowed to expand 
adiabatically in an engine cylinder or in a turbine nozzle to a 
temperature of ioo° F., this expansion will be represented by the 
line EG. GN represents the condensation of the exhaust steam. 
The area NCEG represents the energy in the steam available 
for work in the steam motor. 

PROBLEMS 

1. One pound of water is raised in temperature from 6o° to 90 F. 
What is the increase in entropy? 

2. Dry and saturated steam has a pressure of 100 lbs. per sq. in. absolute. 

(a) Calculate the entropy of the liquid. 

(b) Calculate the entropy of evaporation. 

(c) Calculate the total entropy of the steam. 

(d) Check values in (a), (b), and (c) with the steam tables. 

3. Steam at 150 lbs. per sq. in. absolute has a quality of 0.90. 

(a) What is the entropy of the liquid? 

(b) What is the entropy of evaporation? 

(c) What is the total entropy of the steam? 

4. Steam having a temperature of 300 F. has an entropy of evapora- 
tion of 1. 1 900. What is its quality? 

5. Steam having a pressure of 200 lbs. per sq. in. absolute has a total 
entropy of 1.5400. 

(a) What is the total entropy of dry and saturated steam under the 

given pressure? 

(b) Is the steam wet or dry? 

(c) What is its quality? 

6. Steam having a pressure of 125 lbs. per sq. in. absolute is super- 
heated ioo° F. 

(a) What is the total entropy of dry and saturated steam under the 

given pressure? 

(b) What is the entropy of the superheat? 

(c) What is its total entropy? 

7. Steam having a pressure of 150 lbs. per sq. in. absolute has a total 
entropy of 1.6043. 

(a) What is the total entropy of dry and saturated steam under the 
given pressure? 



PROBLEMS 103 

(b) Is the above steam saturated or superheated? How can you tell? 

(c) How much superheat has the steam? 

8. A boiler generates steam of 0.90 quality at a temperature of 350 F. 
with the feed water admitted at oo° F. What is the increase in entropy? 

9. Plot a temperature-entropy diagram for one pound of water vapor 
for the pressures of 15, 50, 100, 150, 200, 250. The diagram should show the 
liquid line, the 90 per cent quality line, the dry saturated line, and the 50 
and ioo° superheat curves for each of the given pressures. 

10. What is the entropy of steam 92 per cent dry at a pressure of 15 lbs. 
per sq. in. absolute? 

n. With a quality of 0.90, what is the entropy of evaporation of steam 
at a pressure of 25 lbs. per sq. in. absolute? 

12. What is the total entropy of steam 94 per cent dry at a pressure of 
100 lbs. per sq. in. absolute? 



CHAPTER VII 
EXPANSION AND COMPRESSION OF VAPORS 

Vapors, like gases, can be expanded or compressed, but the 
laws governing their thermodynamic relations are more complex. 
In the heat changes of vapors the heat that is required to change 
the state of the substance must be accounted for, in addition to 
that necessary to do the external work and to produce the change 
in temperature of the substance. 

Equation 12, showing the effect of the heat added or abstracted 
during an expansion or a compression, applies equally to vapors 
and gases. 

The external work (equation 1 2) done during the expansion or 
compression of vapor is calculated, as in the case of gases, from 
the area under the expansion or compression curve. Work, 
being a product of pressure and volume, is independent of the 
working medium. 

The internal energy changes (equation 12) involved during an 
expansion or compression of vapors are best measured by dif- 
ferences. The internal energy at the beginning and at the end 
of an expansion or compression may be determined by reference 
to the vapor tables. The values, as obtained from such tables 
(Tables 3, 4, 5, 6 in Appendix), are calculated from a standard 
datum temperature of 32 F., and their difference gives the change 
in internal energy involved in the process in question. 

Thus, for any vapor the following general equation may be 
written: 

Heat added = Internal energy at the end of the expansion or 

compression minus the internal energy at the beginning of the 

expansion or compression plus the heat equivalent of the external 

work done, or 

Q = (h- h) + W. (151) 

104 



EXPANSION OF WET STEAM AT CONSTANT VOLUME 105 

The following problems show the application of equation (151) 
to various types of expansions. 

1. Expansion of Wet Steam at Constant Volume. Assume 
that one pound of steam at a pressure of 15 pounds per square 
inch absolute, and 50 per cent dry, receives heat under constant 
volume raising the pressure to 30 pounds per square inch absolute. 

Find: (a) the volume of the steam after the addition of the 
heat, (b) the quality of the steam, (c) the work done, (d) the 
heat added. 

Solution: (a) Since the volume remains constant the final 
volume of the steam is 

xFsat. = 0.50 X 26.27 = 13.13 cu. ft., 

in which x = the quality of the steam, V = volume of saturated 
steam at 15 pounds per square inch absolute pressure (26.27) 
as obtained from steam tables, in cubic feet per pound. 

(b) From steam tables the volume of one pound of dry and 
saturated steam at the final condition of 30 pounds per square 
inch absolute is 13.74. Since the actual volume is less than that 
of the saturated steam, the steam in its final condition is wet and 
the quality is 

from which 

x = ■ = 0.955 or 95-5 P er cent dry. 

(c) Since the volume is constant the work done is 

W = o. 

(d) The heat added is 

Q = h - h + W 
= h - h + o 
= h - h. 

The internal energy at the final condition is, from equation 
(no), 

h = h+X2 \L 2 * — L \ = h 2 + X2P2 

778 



106 EXPANSION AND COMPRESSION OF VAPORS 

= (218.8 + O.955 X 869.0) 

= 218.8 + 829.9 = 1048.7 B.t.u., 

where fa and p 2 are the heat of the liquid and internal latent 
heat respectively at 30 pounds per square inch absolute pressure. 
The internal energy at the initial condition is 

Ii =\ h + Xipi 

= (181.0 + 0.50 X 896.8) 

= (181.0 + 448.4) = 629.4 B.t.u. 



Then, 



Q = 1048.7 — 629.4 
= 419.3 B.t.u. 



2. Expansion of Superheated Steam at Constant Volume. 

One pound of steam at 130 pounds per square inch absolute pres- 
sure and 50 F. superheat is heated under constant volume to 
a pressure of 180 pounds per square inch absolute. 

Find (a) the final quality of the steam; 
(b) the heat supplied. 

Solution: (a) the initial volume of the steam is found by 
reference to the superheated steam tables to be 3.74 cubic feet. 
Since this equals the final volume of the steam, the final condition 
of the steam is 300 F. superheat, which from the steam tables is 
the condition when steam at 180 pounds per square inch absolute 
pressure has a volume of 3.74 cubic feet. 

(b) The heat supplied equals 

Q = h - h + W 
= J1-/1. 

No values corresponding to the internal energy are available 
in the superheated tables and consequently these must be cal- 
culated. 

*- - ^) to) 

where 



-(■ 



#sup. = the total heat of superheated steam as found in the 
superheat tables. 



EXPANSION AT CONSTANT PRESSURE 107 

P 2 = pressure in pounds per square foot. 

V 2 = volume of the superheated steam, cubic feet per pound. 

Thus, for the conditions of the problem, 

T ( 180 X 144 X 3-74\ 

h = (1353-9 jfe-^) 

= 1 353-9 ~ 124.6 = 1229.3 B.t.u. 

t ( I 3° X x 44 X 3-74-\ 
/ 1= ^ 2I9 . 7 _-3 ^—^±) 

= 1219.7 — 89.5 = 1130.2 B.t.u., 



from which 



Q = 1229.3 — 1130.2 = 99.1 B.t.u. 



3. Expansion at Constant Pressure. One pound of steam 
at a pressure of 150 pounds per square inch absolute and a 
volume of 1.506 cubic feet expands under constant pressure until 
it becomes dry and saturated. 

1. What is the quality at the initial condition? 

The volume of dry and saturated steam at the given pres- 
sure is 3.012 cubic feet per pound. 

The quality then is — — = 0.50. 

3.012 

2. What is the volume of the steam at the final condition? 
The volume of a pound is 3.012 cubic feet since the steam 

is dry and saturated. 

3. What is the work done during the expansion? 

Work = Pi (V 2 - Vi) = 150 X 144 (3.012 - 1.506) = 32,530 
foot-pounds. 

4. How much heat is required? Several methods of reason- 
ing may be used in calculating the values of the heat required 
and the work done during constant pressure expansion or 
compression. 

Q = h - h + W 

h = fa + X2P2 

= 330.2 + I X 780.4 = 1 1 10.6 



108 EXPANSION AND COMPRESSION OF VAPORS 

1 1 = hi + Xipi 

= (330- 2 + 0.50 X 780.4) 
= 720.4 B.t.u. 

The heat equivalent of the external work during the expansion 
is 

^ = ^ i2 C53) 

150 X 144 (1 X 3.012 - 0.50 X 3.012) 

= = A.1.0. 

778 

Substituting in equation (151), 

Q = 1110.6 — 720.4 + 41.8 
= 432.0 B.t.u. 

The amount of heat required in the case of constant pressure 
expansion may be calculated directly: 

Q = (x 2 - Xi) L, (154) 

where x 2 and Xi equal the qualities at the initial and final con- 
ditions of the expansion; L equals the latent heat of dry and 
saturated steam at the given pressure. Thus, for the above 
problem, 

Q = (1.00 — 0.50) 863.2 = 431.6 B.t.u. 

Similarly, the heat supplied during constant pressure expansion 

may be expressed: 

Q = H 2 - H h 

where H 2 and Hi are the total heat of the steam above 32 for 
the initial and final conditions respectively. Thus, for the prob- 
lem in question, 

H 2 = 330.2 + i-oo X 863.2 
Hi = 330.2 + 0.50 X 863.2 

Q = (330.2 + 1.00 X 863.2) - (330.2 + 0.50 X 863.2) 
= 431.6 B.t.u. 

Isothermal Lines for Steam. When the expansion of steam 
occurs at constant pressure as, for example, in the conversion 



ADIABATIC LINES FOR STEAM 109 

of water into steam in a boiler when the engines are working, we 
have isothermal expansion. Isothermal lines for wet steam, 
which consists of a mixture of water and its vapor, are, there- 
fore, straight lines of uniform pressure. On a pressure-volume 
diagram an isothermal line is consequently represented by a 
horizontal line parallel to the axis of abscissas. As steam 
becomes superheated the pressure decreases as the volume 
increases; for highly superheated vapors the isothermal curve 
approaches a rectangular hyperbola. On a T-cf) diagram, the 
isothermal line is represented by a line of constant temperature, 
i.e., by a line parallel to the <£-axis. 

Adiabatic Lines for Steam. Adiabatic lines will have differ- 
ent curvature depending upon the substances used, It will be 
remembered that the values of 7 are different for the various 
gases and therefore the adiabatic line for each of these gases 
would have a different curvature. In the same way the curva- 
ture of adiabatic lines of steam will vary with the quality of the 
steam. Steam which is initially dry, if allowed to expand adia- 
batically, will become wet, the percentage of moisture which it 
will contain depending on the extent to which the expansion is 
carried. Also, on any T-<j> diagram, an adiabatic (isentropic) line 
is represented by a line parallel to the T-axis, i.e., by a line of 
constant entropy. If steam is initially wet and is expanded adia- 
batically, it becomes wetter as a rule.* 

In general, in any expansion, in order to keep steam at the 
same relative dryness as it was initially, while it is doing work, 
some heat must be supplied. If the expansion is adiabatic so 
that no heat is taken in, a part of the steam will be condensed 
and will form very small particles of water suspended in the 
steam, or it will be condensed as a sort of dew upon the surface 
of the enclosing vessel. 

The relation between pressure and temperature as indicated 
by the steam-tables continues throughout an expansion, pro- 
vided the steam is initially dry and saturated or wet. 

* When the percentage of water in steam is very great and the steam is ex- 
panded adiabatically there is a tendency for the steam to become drier. This 
is very evident from an inspection of diagrams like Fig. 25. 



HO EXPANSION AND COMPRESSION OF VAPORS 

Adiabatic Curve for Steam. Whether steam is initially dry 
and saturated or wet, the adiabatic curve may be represented 
by the formula: PV n = constant. The value of the index n 
depends on the initial dryness of the steam. Zeuner has deter- 
mined the following relation: 

n = 1.035 + — (i55) 

10 

Solving this when x = unity (dry and saturated steam) the 
value of n is 1.135, and when x is 0.75, n has the value 1.11.* 

While the work during an adiabatic expansion may be cal- 
culated from the value of n as determined from Zeuner's relation 
(equation 155), the following method is more commonly used. 

Since by definition the heat added or abstracted during an 
adiabatic process is zero, the heat equation becomes, 

Q = I 2 -h + W = 0, 
or W = h - 1 2 (156) 

The work done by an adiabatic process can then be determined 
by the difference between the internal energies at the beginning 
and end of the process. 

In order to calculate the internal energies the qualities must 
be known and these may be found from the entropies. 

Quality of Steam During Adiabatic Expansion. Since in an 
adiabatic expansion no heat transfer takes place, the entropy 
remains constant, and, therefore, on a T-<j> diagram, this condition 
is represented by a straight vertical line as cgf (Fig. 25) or smk. 
Thus for wet steam: 

cf> = 6 + x-, (157) 

where x = the quality or dryness fraction of the steam, <j> = 

* Rankine gave the value of n = V, which obviously from the results given 
is much too low if the steam is at all near the dry and saturated condition. His 
value would be about right for the condition when x = 0.75. In an actual steam 
engine, the expansion of steam has, however, never a close approximation to the 
adiabatic condition, because there is always some heat being transferred to and 
from the steam and the metal of the cylinder and piston. 



QUALITY OF STEAM DURING ADIABATIC EXPANSION III 

the total entropy of saturated steam, and = the entropy of 
the water. 

Since in adiabatic expansion the entropy remains constant, 
the following equation can be written 

0i = 02 (total entropies) 
or 

Bi + xM = e 2 + xM- (i 5 8) 

il 1 2 

Knowing the initial conditions of steam, the quality of the 
steam at any time during adiabatic expansion can be readily 
determined* Thus, suppose the initial pressure of dry satu- 
rated steam to be 100 pounds per square inch absolute, and the 
final pressure after adiabatic expansion 1 7 pounds per square inch. 

From the steam tables we find that the total entropy 0i, for 
dry steam at 100 pounds pressure, is 1.6020; that is 

0i = 1.6020. 

The entropies at 17 pounds pressure are also obtained from the 
tables, 

1.6020 = 0.3229 -f- # 2 14215, 
whence 

#2 = 0.899. 

For a rapid and convenient means of checking the above 
result, the " Total Heat-entropy " diagram in the Appendix can 
be used. From the intersection of the 100-pound pressure line 
and that of unit quality (" saturation line") is dropped a vertical 
line (line of constant entropy = 1.602) to the 17-pound pressure 
line. This latter intersection is found to He on the 0.90 quality 
line. 

Example. One pound of steam having a quality of 0.95 at a 
pressure of 100 pounds per square inch absolute expands adia- 
batically to 15 pounds per square inch absolute. 

1. What is the quality at the final condition? 

The total entropy at the initial condition equals 

+ x- = 0.4743 + 0.95 x 1. 1277 = 1.5456. 



112 EXPANSION AND COMPRESSION OF VAPORS 

The total entropy at the end of the expansion equals 

+ x - = 0.3133 + 1.4416 x. 

Since entropy is constant in adiabatic expansion 
0.3133 + 1.4416 s = i.545 6 > 

from which the final condition is: x = 0.854. 

2. How much work is done during the expansion? 

Since there is no heat added the work done equals the loss in 
internal energy. 

The internal energy at the end of the expansion equals 

h + X2P2 = 181. o + 0.854 X 896.8 = 946.9 B.t.u. 
The internal energy at the initial condition equals 

h + Xipi = 298.3 + 0.95 X 806.6 = 1064.6 B.t.u. 
The work equals 

1064.6 — 946.9 = 1 1 7.7 B.t.u. or 91,576 foot-pounds. 

Poly tropic Expansion {n — 1). 

Example. One pound of steam has a pressure of 100 pounds 
per square inch absolute and a quality of 0.95. It expands 
along an n = 1 curve to 20 pounds per square inch absolute. 

What is the quality at the end of the expansion? 

Solution. The volume of the steam at the initial condition is 

X X F S at. 

x X F S at. = 0.95 X 4.429 = 4.207 cubic feet. 

Obviously, PxV? = P 2 V 2 n 

and since n — 1, 

100 X 4.207 = 20 X V2, 

V 2 = 21.035 cubic feet. 

The volume of dry saturated steam at the end of the expan- 
sion or at 20 pounds per square inch is 20.08. Therefore the 
steam is superheated at end of expansion. How is this known? 



GRAPHICAL DETERMINATION OF QUALITY OF STEAM 113 

Graphical Determination of Quality of Steam by Throttling 
Calorimeter and Total Heat-entropy Diagram. It will be re- 
membered that the throttling calorimeter (pages 77 to 81) de- 
pends for its action upon the fact that the total heat of steam 
which expands without doing work remains the same, the heat in 
excess of that required to keep the steam dry and saturated going 
to superheat the steam. Suppose that steam enters the calorim- 




d.50 



1.60 



1.70 



1.80 



Entropy 
Fig. 28. — Mollier Diagram for Determining Quality of Steam. 

eter at a pressure of 150 pounds per square inch absolute, and is 
throttled down to 1 7 pounds per square inch, the actual tempera- 
ture being 24o ? F. Since the saturation temperature for steam 
at 17 pounds pressure is 219.4, the steam in the calorimeter is 
superheated 240 — 219.4 or 20.6 degrees. In order to find the 
quality of the live steam refer to the " Mollier Diagram " (Appen- 
dix or Fig. 28) and find the intersection of the 20.6 degrees super- 
heat line with the 1 7-pound pressure line. From this point follow 



114 EXPANSION AND COMPRESSION OF VAPORS 

a horizontal line (line of constant total heat) to the left until it 
intersects the 150-pound pressure line. This point of intersec- 
tion is found to lie on the 0.96 quality line. 

Formula 117 (page 80) gives the following result in close 
agreement with the diagram: 

_ 1153.1 + 047 (240 ~ 2194) - 330.2 
Xl ~ 863.2 

= 0.965. 

PROBLEMS 

1. Dry saturated steam at 100 lbs. per sq. in. absolute pressure contained 
in a closed tank is cooled until its pressure drops to 15 lbs. per sq. in. abso- 
lute. What is the final quality and the heat removed from each pound of 
steam? 

2. One pound of steam at 15 lbs. per sq. in. absolute has a volume of 1 2.36 
cu. ft. It is heated under constant volume until the pressure becomes 
50 lbs. per sq. in. absolute, (a) What is the quality before and after the 
heating? (b) How much heat was supplied? 

3. A closed tank containing dry and saturated steam at 15 lbs. per sq. in. 
absolute pressure is submerged in a body of water at a temperature of 59 F. 
What will be the ultimate pressure and quality of steam within the tank? 

4. Prove that for a constant pressure expansion the heat supplied equals 
the difference between the total heats of the vapor at the beginning and at 
the end of the expansion. 

5. One pound of steam at 100 lbs. per sq. in. absolute pressure and 50 per 
cent dry expands at constant pressure. What work is done and what heat 
is required to double the volume? "What is the temperature at the beginning 
and at the end of the expansion? 

6. One pound of steam at a pressure of 100 lbs. per sq. in. absolute 
and 50 per cent dry is expanded isothermally until it is dry and saturated. 
Find the heat supplied and the work done. 

7. One pound of steam at a pressure of 150 lbs. per sq. in. absolute has a 
quality of 0.90. What work is done and what heat is required to double 
its volume at constant pressure? 

8. If steam at 200 lbs. per sq. in. absolute, 95 per cent dry, is caused to 
expand adiabatically to 228 F., what are the properties of this steam at 
the lower point? (That is, final total entropy, entropy of evaporation, 
quality and volume.) 

9. What will be the final total heat of dry saturated steam that is ex- 
panded adiabatically from 150. lbs. per sq. in. absolute down to 10 lbs. 
per sq. in. absolute? 



PROBLEMS 115 

10. Steam having a quality of 0.20 dry is compressed along an adiabatic 
curve from a pressure of 20 lbs. per sq. in. absolute to a pressure correspond- 
ing to a temperature of 293 F. What is the final quality? 

11. Determine the final quality of the steam and find the quantity of 
work performed by 2 lbs. of steam in expanding adiabatically from 250 lbs. 
per sq. in. absolute pressure to 100 lbs. per sq. in. absolute, the steam being 
initially dry and saturated. 

12. One pound of steam at 150 lbs. per sq. in. absolute, and 200 F. 
superheat, expands adiabatically. What is the pressure when the steam 
becomes dry and saturated? What is the work done during the expan- 
sion? 

13. One pound of dry and saturated steam at 15 lbs per. sq. in. absolute 
pressure is compressed adiabatically to 100 lbs. per sq. in. absolute pressure. 
What is the quality at the end of the compression and the negative work 
done? 

14. One pound of steam at 100 lbs. per sq. in. absolute pressure has a 
quality of 0.80. It expands along an n = 1 curve to a pressure of 15 lbs. 
per sq. in. absolute, (a) What is the volume at the beginning and at the 
end of the expansion, (b) the quality at the end of the expansion, (c) the 
work done during the expansion, (d) the heat supplied to produce the expan- 
sion? 

15. One pound of steam at 150 lbs. per sq. in. absolute expands along an 
n = 1 curve to 15 lbs. per sq. in. absolute. The quality of the steam at 
15 pounds is to be dry and saturated, (a) What must be the quality at the 
initial conditions? (b) What work will be done by the expansion? (c) What 
heat must be supplied? 

16. Steam at a pressure of 100 lbs. per sq. in. absolute having a quality 
of 0.50 expands adiabatically to 15 lbs. per sq. in. absolute. What is the 
quality at the end of the expansion? 

17. One pound of steam at a pressure of 100 lbs. per sq. in. absolute has 
a volume of 4 cu. ft. and expands adiabatically to 15 lbs. per sq. in. absolute. 

(a) What is the quality at the initial and final conditions? 

(b) What is the work done during the expansion? 

18. One pound of steam having a pressure of 125 lbs. per sq. in abso- 
lute and volume of 4.17 cu. ft. expands adiabatically to 25 lbs. per sq. in. 
absolute. 

(a) What is the quality at the initial and final conditions? 

(b) What is the work of expansion? 

19. Given the steam as stated in problem 18 but with expansion complete 
at 100 lbs. per sq. in. absolute. What is the quality at this pressure? 

20. What would be the pressure if the steam in problem 18 were expanded 
adiabatically until it became dry and saturated? 



n6 EXPANSION AND COMPRESSION OF VAPORS 

21. One pound of steam at a temperature of 360 F. has a quality of 
0.50, and expands under constant pressure to a volume of 3.4 cu. ft. 

(a) What is the quality at the final condition? 

(b) What is the work of the expansion? 

(c) What heat is required? 

22. Two pounds of steam at a pressure of 100 lbs. per sq. in. absolute 
have a volume of 4 cu. ft., and expand under constant temperature to a 
volume of 8 cu. ft. 

(a) What is the quality at the initial and final conditions? 

(b) What is the work of the expansion? 

(c) How much heat is required? 



CHAPTER VIII 

CYCLES OF HEAT ENGINES USING VAPORS 

Carnot Cycle. The Carnot cycle using a vapor employs the 
same apparatus as was explained on page 40. The cycle is made 
up of two isothermals and two adiabatics, but differs from the 
cycle using gas as the working medium in that the isothermal 
curves are lines of constant pressure. 




Volume 
Fig. 29. — Camot Cycle using Vapors. 

As an illustration, assume that the vapor in the cylinder is in 
the liquid state and at a temperature T\ when the heat is applied. 
The heating process continues until the vapor becomes dry and 
saturated. In order to maintain the temperature constant the 
heat must be supplied at such a rate as to maintain the pressure 
constant. The volume increases during this change from that 
of the specific volume of the liquid to the specific volume of the 
dry saturated vapor at the temperature TV This change is 
represented by the lines ah on the PV and temperature-entropy 

diagrams, Figs. 29 and 30. 

117 



n8 



CYCLES OF HEAT ENGINES USING VAPORS 




The cylinder is then removed from the source of heat and the 
adiabatic expansion is produced. This is represented in Figs. 
29 and 30 by the curves be. 

The isothermal compression curve cd is produced when the 

cylinder is placed in communi- 
cation with the refrigerator or 
condenser and heat is absorbed 
from the vapor. Since the cyl- 
inder contains a saturated va- 
por, the process results in a 
constant pressure curve at tem- 
perature, r 2 . 

The adiabatic compression 
curve da follows when the cyl- 
inder is removed from the re- 
frigerator. The vapor in its 
final condition is reduced to a 
liquid at the temperature 7\. 

The heat added to the cycle 
is that required to produce the 
line ab, Figs. 29 and 30. For the assumed case this is: 

<2i = H b - h a = L h (159) 

where H b = total heat of dry saturated steam at the pressure P b , 
h a = heat of liquid above 32 F. for condition at a. 

The heat rejected from the cycle is that absorbed by the 
refrigerator and for each pound of vapor it is equal to 

Q2 = (h c + XcL c ) — {h d + XdLa). (160) 

From the temperature-entropy diagram, Fig. 30, the heat added 
and heat rejected are 

Qi = T x (fa - <£«), (161) 

ft = T 2 (fa - 0d). (162) 

Since (</> c — </><*) = (<£& — <t>a), 

Q 2 = T 2 (</>» - 4> a ). (163) 



Entropy-^ 

Fig. 30. — Temperature-entropy Diagram 
of Carnot Cycle using Vapors. 



THE RANKINE CYCLE 119 

The work of the cycle may be determined from the algebraic 
sums of the work under the individual curves, but can be more 
simply calculated from the temperature-entropy diagram, Fig. 
30. The area abed is proportional to the work and since it is a 
rectangle, 

work of cycle = (T 2 — 7\) (<fr> — <£ a ). 

From equations (161) and (163), the efficiency of the cycle is 

77 _ (T2 — Ti) ((f> b — a ) 



T\ (</>& — <f>a) 

T 2 -T x 



(164) 



Equation (164) shows that the efficiency of Carnot's cycle is 
not affected by the character of the working substance and is 
dependent only upon its initial and final temperatures. 

The Rankine Cycle.* The Carnot cycle gives the maximum 
efficiency obtainable for a heat engine operating between given 
limits of temperature. In order that a steam engine may work on 
a Carnot cycle, the steam must be evaporated in the cylinder 
instead of in a separate boiler, and condensed in the cylinder 
instead of being rejected to the air or to a separate condenser. 
Such conditions are obviously impracticable, and it has, therefore, 
been found necessary to adopt some other cycle which conforms 
more with practical conditions. The most efficient practical 
steam cycle, and the one which has been adopted as the standard 
with which the efficiency of all steam engines may be compared 
is the Rankine Cycle. The pressure- volume diagram of this cycle 
is shown in Fig. 31. Steam is admitted at constant pressure and 
temperature along ab to a cylinder without clearance. At b cut- 
off occurs, and the steam expands adiabatically from b to c, some 
of i condensing during the process. The steam is then dis- 
charged at constant pressure and temperature along the back 
pressure line cd. Line da represents the rise in temperature and 

* Also known as the Clausius Cycle, having been published simultaneously 
and independently by Clausius. 



120 



CYCLES OF HEAT ENGINES USING VAPORS 



pressure at constant volume when the condensed steam is con- 
verted into a vapor. 
The four stages of the Rankine cycle may be stated as follows: 

(i) Feed water raised from temperature of exhaust to tem- 
perature of admission steam. (Line da.) 

(2) Evaporation at constant admission temperature. (Line 
ab.) 

(3) Adiabatic expansion down to back pressure. (Line be.) 

(4) Rejection of steam at the constant temperature corre- 
sponding to the back pressure. (Line cd.) 




Volume 

Fig. 31. — Indicator Diagram of Ideal Rankine Cycle. 

The net work done in the cycle, assuming one pound of wet 
steam, is calculated as follows: 

1. External work of evaporation (in B.t.u.), 

(W») = tIs (PiFi*i). (165) 

2. Loss in internal energy, 

(W bc ) = fa + Km — (fa + X2P2) B.t.u. (166) 

3. External work of evaporation at temperature of exhaust 

(B.t.u.), 

W dc = - Thf (PcVc) = - T^8 (P2V 2 X 2 ). (167) 

4. W da - O. (168) 



THE RANKINE CYCLE 121 

Adding, 
Net work of cycle, 

W = -ffaPiViXi + hi + xipi - h- x 2 p 2 

— TT 8" -P2 V 2 X 2 , 

but 

•yyg P^V\ + pi — -£1. 
Therefore, 

Tf = h + X1Z1 - (fe + x 2 L 2 ) (B.t.u.). (169) 

Thus the net work of the Rankine cycle is equal to the dif- 
ference between the total heat of the steam admitted and the 
total heat of the steam exhausted. This statement applies 
whether the steam is initially wet, dry or superheated. 

The heat added during the cycle is 

Qi = h a + -XbLj, — h d . (170) 

The heat rejected during the cycle is 

Q 2 = XcLc. (171) 

The efficiency of the Rankine cycle is 

E __ Qi— Q2 = h a + x b L b — ha — XcLc t 
Qi h a + x^ — h d 

Since h a = h b and h c = h d , 

77 f^b ~~ X^L,^ He XcL/c / \ 

rt b -j- XfrL b lie 

Fig. 32 is the T-<j> diagram for a Rankine cycle using dry 
saturated steam. The letters abed refer to the corresponding 
points in the pressure- volume diagram Fig. 31. The net work 
of the cycle is the area B + C, which is the difference between 
the total heats at admission and exhaust. The heat added per 
cycle is represented by the areas A + B + C + D and the 

B + C 
Thermal efficiency = A + B + Q + D (173) 

hi + X\Li — h 2 — x 2 L 2 ( \ 

= j — ; j 7 ^74; 

hi + X1L1 — rh 

* The final quality can be determined by equation (158). 



122 



CYCLES OF HEAT ENGINES USING VAPORS 



fe in the denominator must always be subtracted from hi + X\L\ 
(the total heat above 3 2° F.), in order to give the total heat above 
the temperature of feed water, which in engine tests is always 
assumed for the purpose of comparison to be the same as the 
exhaust temperature. 




Entropy— 
Fig. 32. — Temperature-entropy Diagram of Rankine Cycle. 

Example. One pound of steam at a pressure of 160 pounds per 
square inch absolute and quality of 0.95 performs a Rankine cycle 
exhausting at 5 pounds per square inch absolute. 

1. What is the quality of the exhaust? 
The total entropy at the initial condition 

= 0.5208 + 0.95 X 1.0431. 

The total entropy at the exhaust 

= 0.2348 + 1.608 x. 

Then 0.5208 + 0.95 X 1.0431 = 0.2348 + 1.6084 #. 
From which x = 0.794. 

2. What is the net work of the cycle? 

Work = Hi - H 2 = 335.6 + 0.95 X 858.8 - 130.1 - 0.794 

x 1000.3 = 227.2 B.t.u. 

3. What is the efficiency of the cycle? 



THE RANKINE CYCLE 



123 



Efficiency = 



227.2 



335.6 + 0.95x858.8-130.1 



= 0.222 or 22.2 per cent. 



In the actual steam engine cycle it is impractical to expand the 
steam down to the back-pressure line. 

The Rankine cycle for the actual steam engine is similar to 
that described, except that the adiabatic expansion terminates at 
a pressure higher than that of the back-pressure, that is, the 
expansion is incomplete. 

The pressure-volume and T-4> diagrams of this modified 
Rankine cycle using dry saturated steam are shown in Figs. 33 
and 34. The cylinder is without clearance. Steam is admitted 




Volume 
Fig. 33. — Rankine Cycle, Incomplete Expansion. 

at constant pressure and temperature along ab. Cut-off occurs 
at point b and the steam expands adiabatically to the terminal 
pressure c. Part of the steam is discharged at constant volume 
cd. The remainder is exhausted during the back-pressure stroke 
de. Line ea represents the rise in temperature of the feed water 
from T 2 to 2\. 

The net work of the cycle can best be calculated by dividing 
Fig* 33 into the two areas abed and e'ede. The area abed is that 
of the theoretical or ideal Rankine cycle, while dede is that of the 
rectangle. The net work of the cycle is equal to the sum of these 
two areas. 



124 



CYCLES OF HEAT ENGINES USING VAPORS 



The heat equivalent of the area abed is 

hb + XbLb — h c — XcL c . 
The area e'ede is 

Tb (Pc - Pa) (Va - o). 

The total heat equivalent of the work of the cycle is 

(Pc - Pa) Va 



h + XbLb — h c — XcLc + 

The heat added to the cycle is 

hb + XbLb — ha. 
The efficiency of the cycle is 



778 



hbXbLb — h c —XcLc + 



E = 



{Pc - Pa) Va 
778 * 



hb + XbLb — ha 



(i75) 
(176) 

(i77) 
(178) 

(i79) 



In the T-<j> diagram, Fig. 34, the letters abede refer to 
corresponding points in the pressure-volume diagram (Fig. 
33). The net work of the cycle is represented by the area B 

+ C. The heat added to the 
cycle is represented by the area 
A + B + C + D. 

The incomplete Rankine cy- 
cle (Figs. 33 and 34) is less 
efficient than the ideal cycle 
(Figs. 31 and 32), because of 
failure to expand the steam 
completely. This loss is repre- 
sented in Figs. 33 and 34 graphi- 
cally by the area cdf. This 
cycle is sometimes used as a 
standard in preference to the 
ideal Rankine cycle when the 
efficiencies of engines are com- 
pared. 

Example. One pound of steam at a pressure of 160 pounds 
per square inch absolute and quality of 0.95 goes through a 




Entr.opy-0 
Fig. 34. — Rankine Cycle, Incomplete 
Expansion. 



THE PRACTICAL OR ACTUAL STEAM ENGINE CYCLE 125 

Rankine cycle. The terminal pressure is 15 pounds per square 
inch absolute and the exhaust pressure 5 pounds per square inch 
absolute. What is (a) the work of the cycle, (b) the heat added 
to the cycle, and (c) the efficiency of the cycle? 

Solution. The total entropy of the steam at the initial con- 
dition is 

0.5208 + 0.95 X 1.0431. 

The total entropy of the steam at the terminal pressure is 

0.3133 + x (1.4416). 

Then 0.5208 + 0.95 X 1.0431 = 0.3133 + x (1.4416) 
from which the quality at the terminal pressure is 

x = 0.831. 

The volume of the steam at the terminal pressure is 
0.831 X 26.27 = 21.83. 

The total heat of the steam at the initial condition is 

335.6 + 0.95 X 858.8 = 1151.5. 
The total heat of the steam at the terminal pressure is 

181.0 + 0.831 X 969.7 = 986.8. 
The work of the cycle is then (in terms of heat units), 

Trr on o . ( T 5 ~ 5) X X 44 X 21.83 

w = 1151.5 - 986.8 + ^-2 — D/ r* * = 205.1. 

778 

The heat added to the cycle is 

335.6 + 0.95 x 858.8 — 130.1 = 1021.4 B.t.u. 

The efficiency of the cycle is 

= 20.08 per cent. 



1 02 1. 4 

The Practical or Actual Steam Engine Cycle. In the steam 
engine designed for practical operation it is impossible to expand 
the steam down to the back-pressure line; and, furthermore, it 



126 



CYCLES OF HEAT ENGINES USING VAPORS 



is evident that some mechanical clearance must be provided. 
The result is that in the indicator diagram from the actual steam 
engine, we have to deal with a clearance volume, and both incom- 
plete expansion and incomplete compression as shown in Fig. 
35. In order to calculate the theoretical efficiency of this prac- 



9 












b 


c\ 




h 




a 

















% 




[ 






83 




\ 






<o 




\ 






Si 




\ 






Ph 




\ 






i 

3 




\ 


Sl_ 


- '!»# 






Xf 









Volume 
Fig. -35. — Indicator Diagram of Practical Engine Cycle. 

tical cycle, it is necessary to assume that the expansion line cd 
and the compression line fa are adiabatic. Knowing then the 
cylinder feed of steam per stroke and the pressure and volume 
relations as determined from the indicator diagram, one can 
calculate the theoretical thermal efficiency by obtaining the net 
area of the diagram (expressed in B.t.u.) and dividing by the 
heat supplied per cycle. In order to obtain the net area of the 
diagram, the latter may be divided up into several simple parts 
as follows: 

gcdi — area equivalent to a Rankine cycle, 
idej — a rectangle, 

hafj — area equivalent to a Rankine cycle (negative), 
gbah — a rectangle (negative). 

The actual efficiency of the steam engine is usually determined 
by dividing the heat equivalent to a horse power by the heat in 
the steam required to produce a horse power. Since one horse 



AN ENGINE USING STEAM WITHOUT EXPANSION 1 27 



power per hour is equal to ^M^L or 2545 B.t.u., the actual 

efficiency of a steam engine is 

E = 



2545 



WR(H-h) (l8o) 

In equation (180) WR is the water rate or the steam consump- 
tion per horse power per hour, H is the total heat in the steam 
at the initial pressure and quality as it enters the engine, h is the 
heat in the feed water corresponding to the exhaust pressure. 

Efficiency of an Engine Using Steam Without Expansion. In 
the early history of the steam engine, nothing was known about 
the " expansive " power of steam. Up to the time of Watt in 
all steam engines the steam was admitted at full boiler pressure 
at the beginning of every stroke and the steam at that pressure 

carried the piston forward 
to the end of the stroke 
without any diminution of 
pressure. Under these cir- 
cumstances the volume of 
steam used at each stroke 
at boiler pressure is equal to 
the volume swept through 
b}r'the piston. 

An indicator diagram rep- 
resenting the use of steam 
in an engine without expan- 
sion is shown in Fig. 36. 
This diagram represents 
steam being taken into the engine cylinder at 1 at the boiler 
pressure. It forces the piston out to the point 2 when the 
exhaust opens and the pressure drops rapidly from 2 to 3. 
On the back stroke from 3 to 4 steam is forced out of the 
cylinder into the exhaust pipe. At 4 the pressure rises rapidly 
to that at 1 due to the rapid admission of fresh steam into 
the cylinder. In this case the thermal efficiency (E) is repre- 
sented by 




— s>- Volume 
Fig. 36. — Indicator Diagram of an Engine 
using Steam without Expansion. 



128 CYCLES OF HEAT ENGINES USING VAPORS 

E = work done = (P x - P 4 ) (F 2 - V x ) ^ . g , 

heat taken in 778 (x^Li + h — fe) ? 

where the denominator represents the amount of heat taken in, 
with the feedwater at temperature of exhaust, t 2 . In actual 
practice the efficiency of engines using steam without expansion 
is about 0.06 to 0.07, when the temperature of condensation is 
about ioo° F. When steam is used in an engine without ex- 
pansion and also without the use of a condenser the value of 
this efficiency is still lower. It will be observed that under the 
most favorable conditions obtainable the efficiency of an engine 
without expansion cannot be made under normal conditions to 
exceed about 7 per cent. 

In the actual Newcomen steam engines the efficiency was very 
much lower than any of the values given because at every stroke 
of the piston a very much larger amount of steam had to be 
taken in than that corresponding to the volume swept through 
by the piston on account of a considerable quantity of steam 
condensing on the walls of the cylinder. 

Adiabatic Expansion and Available Energy. A practical ex- 
ample as to how the temperature-entropy diagram can be used 
to show how much work can be obtained by a theoretically per- 
fect engine from the adiabatic expansion of a pound of steam 
will now be given. When steam expands adiabatically — with- 
out a gain or loss of heat by conduction — its temperature falls. 
Remembering that areas in the temperature-entropy diagram 
represent quantities of heat and that in this expansion there is 
no exchange of heat, it is obvious that the area under a curve of 
adiabatic expansion must be zero ; this condition can be satisfied 
only by a vertical line which is a line of constant entropy. 

The work done during an adiabatic process, while it cannot be 
obtained from a "heat diagram," can very readily be determined 
from the area under the adiabatic curve of a pressure-volume 
diagram, or better still by the use of steam- tables as follows: In 
an adiabatic expansion the amount of work done is the mechani- 
cal equivalent of the loss in internal energy as explained in Chap- 
ter III. Therefore, it is only necessary to determine the internal 



ADIABATIC EXPANSION AND AVAILABLE ENERGY 129 

energy of the steam at the beginning and end of the adiabatic 
expansion. 

h = h x + Xipi, (182) 

I 2 = h 2 + x 2 p 2 . (183) 

Work during adiabatic expansion = loss in internal energy 

W = (hi + Xipi — Jh — X2P2) 778 (foot-pounds). (184) 

Fig- 37 is a temperature-entropy diagram representing dry 
saturated steam which is expanded adiabatically from an ini- 
tial temperature Z\ (corresponding to a pressure Pi) to a lower 
final temperature T 2 (corresponding to a pressure P 2 ), The 







C / T, P, \ 


E 






JlBB 








"J 


^^^^^^^^^ 


G \ 


g 


1 















F 




F' 







& 



Entropy a -r 2 

Fig. 37. — Temperature-entropy Diagram for Dry Saturated Steam 
Expanded Adiabatically. 

initial and final conditions of total heat (H) and entropy (<p) 
are represented by the same subscripts 1 and 2. The available 
energy or the work that can be done by a perfect engine under 
these conditions is the area NCEG. It is now desired to obtain 
a simple equation expressing this available energy E a in terms 
of total heat, absolute temperature and entropy. 

E x = area OBNCEF, 

H 2 = area OBNG / F / , 

E a = area (OBNCEF + FGG'F') - OBNG'F', 

E a = H 1 -H 2 + (c P 2- <l>i) 2V* (185) 

* It should be observed that this form is for the case where the steam is 
initially dry and saturated. For the case of superheated steam a slightly differ- 
ent form is required. 



130 CYCLES OF HEAT ENGINES USING VAPORS 

An application of this equation will be made to determine the 
heat energy available from the adiabatic expansion of a pound 
of dry saturated steam from an initial pressure of 165 pounds 
per square inch absolute to a final pressure of 15 pounds per 
square inch absolute. 

Example. Pi = 165 7\ = 826 degrees, from steam tables. 

P 2 = IS T 2 = 673.0 

#1 = 1195.0B.tu. " 

#2 = 1150.7 B.t.u. " 

01 = I-56I5 
<h = 1.7549 

Substituting these values in equation (185), we have 

E a = 1195.0 - 1150.7 + (i.7549 - 1-5615) 673 = 174.5 B.t.u. 
per pound of steam. 

Now if in a suitable piece of apparatus like a steam turbine 
nozzle, all this energy that is theoretically available could be 
changed into velocity, then we have by the well-known formula 
in mechanics, for unit mass,* 

V 2 

— = E a (foot-pounds) = E a (B.t.u.) X 778, 

2 g 

V = V778 X 2 gE a = 223.8 VE a , (l86) 

where V is the velocity of the jet and g is the acceleration due to 
gravity (32.2), both in feet per second. 

Solving then for the theoretical velocity obtainable from the 
available energy we obtain the following: 

V = 223.8 V174.5 = 223.8 X 13.22 = 2956 feet per second. 

The important condition assumed as the basis for determining 
equation (185), that the steam is initially dry and saturated, must 
not be overlooked in its application. There are, therefore, two 
other cases to be considered: 

(1) When the steam is initially wet, 

(2) When the steam is initially superheated. 

* See Church's Mechanics of Engineering, page 672, or Jameson's Applied Me- 
chanics and Mechanical Engineering, vol. I, page 47. 



AVAILABLE ENERGY OF WET STEAM 



131 



Available Energy of Wet Steam. The case of initially wet 
steam is easily treated in the same way as dry and saturated 
steam. Fig. 38 is an example of the case in hand. At the ini- 
tial pressure P\ the total heat of a pound of wet steam (hi + X1L1) 
is represented in this diagram by the area OBNCE^F". The 
initial quality of the steam (xi) is represented by the ratio of the 

CE" 

The available energy from adiabatic expansion from 



lines 



CE 



the initial temperature Ti (corresponding to the pressure Pi) to 




Entropy (p x ^ 

Fig. 38. — Temperature-entropy Diagram of Wet Steam Expanded Adiabatically. 

the final temperature T 2 (corresponding to the pressure P 2 ) is the 
area NCE"G". If we call this available energy E aw , we have by 
manipulation of the areas, 

Eaw = area OBNCEF + FGG'F' - OBNG'F' - G"E"EG, 
E aw = H 1 -H 2 + (<p 2 - fa) T 2 - (0! -0,) {Ti - T 2 ) * (187) 
or, 

E aw = H l -H 2 + (0 2 - 0i) T 2 - ^ (1 - Xi) (Ti - T 2 ). (188) 

J- 1 

The velocity corresponding to this energy is found by substi- 
tution in equation (186), just as for the case when the steam was 
initially dry and saturated. 



■t 1 



<t>X = Xi 7jT + 01, 01 — #6 = -J (i — #l). 
1 1 ll 



132 



CYCLES OF HEAT ENGINES USING VAPORS 



Example. Calculations for the velocity resulting from adia- 
batic expansion for the same conditions given in the preceding 
example, except that the steam is initially 5 per cent wet, are 
given below. 

Pi = 165 lbs. absolute. Ti = 826 degrees from tables 
P 2 = 15 lbs. absolute. T 2 = 673.0 degrees 

Hi = 1195.0 B.t.u. 

H 2 = 1150.7 " 
0i = 1.5615 

02 = 1-7549 

Li = 856.8 B.t.u. 

Xi = 1. 00 — 0.05 = 0.95 



E c 



(1195.0 - 1150.7) + (i.7549 - 1-5615) 673 - 



856.8 
826 



X (i - 0.95) (826 - 673) = 44.3 + 130.2 - 7.93 
Eaw = 166.5 B.t.u. per pound of wet steam 

V = 223.8 VeZ = 223.8 x V166.5 

= 223.8 X 12.9 = 2886 feet per second. 

This result can be checked very quickly by the "total heat- 
entropy" or "Mollier" diagram (Appendix). The intersection 
of the 0.95 quality line and 165 pounds pressure line is found to 
lie on the n 52 B.t.u. total heat line. Since the expansion is 
adiabatic, the entropy remains constant. Therefore, following 
the vertical or constant entropy line (entropy = about 1.507) 
down to its intersection with the 1 5 pounds pressure line, we find 
that the total heat at the end of adiabatic expansion is 985 B.t.u 

and 

1 152 — 985 = 167 B.t.u. available energy, as above. 

If the steam were initially superheated, the available energy 
during adiabatic expansion could be obtained in the same way by 
means of the diagram. 

Available Energy of Superheated Steam. The amount of 
energy that becomes available in the adiabatic expansion of 
superheated steam is very easily expressed with the help of Fig. 
39. Two conditions after expansion must be considered: 



AVAILABLE ENERGY OF SUPERHEATED STEAM 



1 S3 



(i) When the steam in the final condition is superheated, 
(2) When the steam in the final condition is wet (or dry 
saturated). 

Using Fig. 39 with the notation as before except E as is the avail- 
able energy from the adiabatic expansion of steam initially super- 
heated in B.t.u. per pound, S and H s are respectively the total 
entropy and the total heat of the superheated steam at the 

T 3 H3 



l* 



Ti 




Entropy — <j> 
Fig. 39. — Temperature-entropy Diagram for Superheated Sxeam. 

initial condition, then from the diagram, when the steam is 
wet at the final condition, • 

E fls = H S -H 2 + (0 2 - <j> s ) T 2 . (189) 

When the steam is superheated at the final condition, 

E as = H s - H 2 ; - (0, - 2 O TV. (190) 

It will be observed that these equations (189) and (190) are the 
same in form as (184), and that equation (190) differs only in hav- 



134 CYCLES OF HEAT ENGINES USING VAPORS 

ing the terms H s and fa in the place of Hi and fa. In other 
words equation (184) can be used for superheated steam if the 
total heat and entropy are read from the steam tables for the 
required degrees of initial superheat. 

The following examples illustrate the simplicity of calcula- 
tions with these equations: 

Example 1. Steam at 150 pounds per square inch absolute 
pressure and 300 F. superheat is expanded adiabatically to 1 
pound per square inch absolute pressure. How much energy in 
B.t.u. per pound is made available for doing work? 

Solution. H s = 1348.8 B.t.u. per pound, 
H 2 = 1 103.6 " 
fa = 1.980, 

4>> = i-73 2 > 
. T 2 = 559-6° F, 

Eas = I348.8 - IIO3.6 + (1.980 - I.732) 559.6 

= 383.9 B.t.u. per pound. 

The result above may be checked with the total heat-entropy 
chart (Appendix) and obtain thus (1349 — 967) or 382 B.t.u. 
per pound. 

Example 2. Data the same as in preceding example except 
that the final pressure is now 35 pounds per square inch absolute. 
(Final condition of steam is superheated.) Calculate Eas. 

Solution. H s = 1348.8 B.t.u. per pound, 
HJ = 1 166.8 " " " 
fa = 1.732, 
fa' = 1.6868, 
TV = 718.9° F., 

Eas = 1348.8 - 1166.8 - (1.7320 - 1.6868) 718.9 
= 149.5 B.t.u. per pound. 

Application of Temperature-entropy Diagram to Analysis of 
Steam Engine. The working conditions of a steam engine, as 
stated before, can be shown not only by the indicator card, but 
also by the employment of what is known as a "temperature- 



APPLICATION OF TEMPERATURE-ENTROPY DIAGRAM 135 

entropy" diagram. This diagram represents graphically the 
amount of heat actually transformed into work, and in addition 
the distribution of losses, in the steam engine. 

For illustration, a card was taken from a Corliss steam engine 
having a cylinder volume of 1.325 cubic feet, with a clearance 
volume of 7.74 per cent, or 0.103 cubic feet; the weight of steam 
in pounds per stroke (cylinder feed plus clearance) was 0.14664 
pounds. Barometer registered atmospheric condition as 14.5 
pounds per square inch. The scale of the indicator spring used 
in getting the card was 80 pounds to the inch. Steam chest 
pressure was taken as 153 pounds per square inch (absolute), 
and a calorimeter determination showed the steam to be dry 
and saturated. 

The preliminary work in transferring the indicator card to a 
T-<f> diagram, consists first in preparing the indicator card as 
follows: It was divided into horizontal strips at pressure intervals 
of 10 pounds with the absolute zero line taken as a reference; 
this line was laid off 14.5 pounds below atmospheric conditions. 
(See Fig. 40.) For reference, the saturation curve was drawn. 
Knowing the weight of steam consumed per stroke and the 
specific volume of the steam (from the Steam Tables), for 
various pressures taken from the card, the corresponding actual 
volumes could be obtained; this operation is, merely, weight of 
steam per stroke multiplied by specific volume for some pressure 
(0.14664 X column 5 in the table below), the resulting value be- 
ing the volume in cubic feet for that condition. These pressures 
and volumes were plotted on the card and the points joined, re- 
sulting in the saturation curve, 2 // -6"-8 // -o/ / . 

The next step consisted in constructing the " transformation 
table " with the columns headed as shown. All the condensing 
and evaporation processes are assumed to take place in the cylin- 
der and the T-cf> diagram is then worked up for a total weight of 
one pound of steam as is customary. Column 1 shows the re- 
spective point numbers that were noted on the card; column 2, 
the absolute pressures for such points; column 3, the corre- 
sponding temperatures for such pressures; column 4, the vol- 



136 



CYCLES OF HEAT ENGINES USING VAPORS 




ENTROPY DIAGRAM 




1.0 

Entropy 

Fig. 40. — Temperature-entropy Diagram of Actual Steam Engine Indicator 

Diagram. 



APPLICATION OF TEMPERATURE-ENTROPY DIAGRAM 137 



ume in cubic feet up to the particular point measured from the 
reference line of volumes; column 5, the specific volume of a 
pound of dry and saturated steam at the particular pressure 
(Steam Tables); column 6, the volume of actual steam per 
pound, obtained by dividing the volumes in column 4 by 0.14664 
pound (total weight of steam in cylinder per stroke) ; column 7, 
the dryness fraction "x," found by dividing column 6 by column 

5 ; column 8 is the entropy of evaporation ( — J f or particular con- 
ditions (Steam Tables); column 9 is the product of column 7 
and column 8; column 10 is the entropy of the liquid at various 
conditions as found in the Steam Tables; column 11 is the sum 
of column 9 and column 10, giving the total entropy. 

TRANSFORMATION TABLE 



I 


2 


3 


4 


5 


6 


7 


8 


9 


IO 


11 


u 

0> 

.a 
6 

a 


u 

<a 

a, 

tA w 

,0.0 

— ' a) 

w .4. 

<l> xn 
u 


1-4 


O.S 

•^ 

s.s 

>8 


in 

."gto 


"c3 

CJ»- ' _ 
C8 «... 

O ft** -1 

8 6 3 

O w 

> 




CO 

C O 


.2 

P. njH 

O U \ 




'B 

a"— 



u 

a 


a 
2 


H 


1 
2 
5 
9 
12 

is 
18 

20 


145 
140 
I20 

48 
20 

8 

8 

3° 


356 

353 
34i 
279 
228 
183 

183 
250 


0.1050 
0.2250 
0.4230 
0.8500 
1 .4225 
0.9758 
0.3500 
0.1825 


3 

3 

3 

8 

20 

47 

47 

13 


11 
22 
73 
84 
08 
27 
27 
74 


0.716 

1-536 

2 .89O 

5-8io 
9.700 
6.660 
2.390 
1.245 


0.2302 
0.4770 
0.7740 
0.6580 
0.4830 
0.1410 
. 0506 
0.0907 


I .0612 
I .0675 

1-0954 
1-2536 

1-3965 
1-5380 
1-5380 

I-33H 


O . 2440 
O . 5090 
0.8475 
O.825O 
O.674O 
0.2l68 
O.O778 
O . I 208 


0.5107 
0.5072 
0.4919 
0.4077 

0-3355 
0.2673 
0.2637 
0.3680 


0.7547 
1 .0162 

1-3394 
1.2327 
1.0095 
0.4841 

0-345I 
0.4888 



Above table is employed for transferring the P-V diagram to the T-<t> diagram. 

After this table was completed, columns 3 and 1 1 were plotted 
(Fig. 40). Convenient scales were selected, the ordinates as tem- 
peratures and the abscissas as entropies. The various points, 
properly designated, were connected as shown on the T-<f> dia- 
gram, the closed diagram resulting. This area shows the amount 
of heat actually transformed into work. This diagram is the 
actual temperature-entropy diagram for the card taken and may 



138 CYCLES OF HEAT ENGINES USING VAPORS 

• 

be superimposed upon the Rankine cycle diagram in order to 
determine the amount and distribution of heat losses. 

The water line, A-A', and the dry steam line, C-C , were 
drawn directly by the aid of Steam Table data, i.e., the entropy 
of the liquid and the entropy of the steam taken at various 
temperatures, and plotted accordingly. 

Before the figure could be studied to any extent, the theoreti- 
cal (Rankine) diagram had to be plotted, assuming that the steam 
reaches cut-off under steam-chest conditions; that it then ex- 
pands adiabatically down to back pressure and finally exhausts 
at constant pressure to the end of the stroke without compres- 
sion. This diagram is marked, A-C-E-H, on the T-4> plane. 
The steam-chest pressure of 153 pounds per square inch absolute 
fixes the point, C, when the temperature line cuts the steam line, 
C-C . The rest of the cycle is self-evident. 

Referring to the T-<f> diagram, Fig. 40, H-A-C-E is the Ran- 
kine cycle with no clearance for one pound of working fluid. 
The amount of heat supplied is shown by the area, Mi-H-A-C-N, 
and of this quantity, the area M\-H-E-N * would be lost in 
the exhaust while the remainder, H-A-C-E, would go into 
work. This is theoretical, but in practice there are losses, and 
for that reason, the Rankine cycle is used merely for comparison 
with the actual card as taken from a test. The enclosed irregu- 
lar area, 1-2-3 • • • 22-2 3> * s the amount of heat going into 
actual work. By observation, it is evident that a big area re- 
mains; this must represent losses of some sort or other. That 
quantity of work represented by the area, i-5-5 / -i / , is lost on 
account of wire-drawing; the area $'-C-D-F shows a loss due 
to initial condensation; the loss due to early release is shown 
by the area F-12-14-F' for the real card, and by D-G-E for the 
modified Rankine cycle (such a loss, in other words, is due to 
incomplete expansion); that quantity represented by 22-B-i f -i 
is lost on account of incomplete compression, and H-A-B-iS is 

* The areas Mi HACN and Mi HEN should have added to them, the rectangu- 
lar area between MiN, and the absolute zero line, which is not shown on the 
diagram (Fig. 40). 



APPLICATION OF TEMPERATURE-ENTROPY DIAGRAM 139 




1.6 2.0 

Entropy 

Fig. 41. — Temperature (absolute) -entropy Diagram of Actual Steam Engine 

Indicator Diagram. 



140 CYCLES OF HEAT ENGINES USING VAPORS 

the loss due to clearance. The expansion line from 5 on to 9 in- 
dicates that there is a loss of heat to the cylinder walls, causing 
a decrease of entropy; from 9 on to 10, re-evaporation is taking 
place (showing a gain of entropy). 

All of the heat losses are not necessarily due to the transfer 
of heat to or from the steam, as there may be some loss of steam 
due to leakage. In general, however, the T-<f> diagram is satis- 
factory in showing heat losses. 

Fig. 41 was constructed for the purpose of showing how the 
actual thermal efficiency and the theoretical thermal efficiency 
(based on the Rankine cycle) can be obtained from the T-<f> 
diagram. The letters in Fig. 40 refer to the same points as 
in Fig. 41, the only difference between the two diagrams being 
the addition of the absolute zero temperature line to Fig. 41. 



Thermal efficiency = 
Actual thermal efficiency = 



Work done 
Heat added 
Shaded area W 
M^H-A-C-N'' 



= —7^ — . (Planimeter) 
20.00 sq. in. 

= 0.096, or 9.6 per cent.* 

Rankine cycle efficiency (theoretical thermal efficiency) 

H-A-C-E 
" Mi'-H-A-C-N'' 

= ~.* " (by planimeter) 
20.00 

= 0.203, or 20.3 per cent. 

Combined Indicator Card of Compound Engine. The method 
of constructing indicator diagrams to a common scale of vol- 

* This value may be regarded as the actual thermal efficiency for one stroke. 

inasmuch as in the succeeding strokes the "cushion" steam will be used over and 

over again and hence will not constitute a heat loss. The actual thermal efficiency 

W 
of a steam engine under running conditions would then be given by , ^rff 

where 18 BCE is a Rankine cycle with clearance and complete compression. 



COMBINED INDICATOR CARD OF COMPOUND ENGINE 141 

ume and pressure shows where the losses peculiar to a compound 
steam engine occur, and, to the same scale, the relative work 
areas. 

As the first step divide the length of the original indicator dia- 
grams into any number of equal parts (Fig. 42), erecting per- 
pendiculars at the points of division. 

In constructing a combined card, select a scale of absolute 
pressure for the ordinates and a scale of volumes in cubic feet 



145.6* 



S7.ir 




Zero Line 



Fig. 42. — High and Low Pressure Indicator Diagram of Compound 

Steam Engine. 

for the abscissas. To the scale adopted draw in the atmos- 
pheric pressure (" at.") line, see Fig. 43. 

Lay off the low-pressure clearance volume on the x-axis to 
the scale selected. In like manner lay off the piston displace- 
ment of the low-pressure cylinder and divide this length into the 
same number of equal parts as the original indicator diagrams 
were divided. From the original low-pressure card (Fig. 42), de- 
termine the pressures at the points of intersection of the perpen- 
diculars erected above the line of zero pressure, taking care 
that the proper indicator-spring scale is used. Lay off these 
pressures along the ordinates (Fig. 43), connect the points and 
the result will be the low-pressure diagram transferred to the 
new volume and pressure scales. 



142 CYCLES OF HEAT ENGINES USING VAPORS 

The high-pressure diagram is transferred to the new volume 
and pressure scale by exactly the same means as described for 
the low-pressure diagram. 

The saturation curve is next drawn. This curve represents 
the curve of expansion which would be obtained if all the steam 
in the cylinder was dry and saturated. It is very probable 
that the saturation curve for each cylinder will not be continuous 
since the weight of the cushion steam in the low-pressure cylin- 
der is usually not the same as that in the high-pressure cylinder. 
The saturation curve would be continuous for the two cards only 
when the weight of cushion steam in the high-pressure and low- 
pressure cylinders is the same (assuming no leakage or other 
losses). 

On the assumption that the steam caught in the clearance 
spaces at the beginning of compression is dry and saturated, the 
weight of the cushion steam can be calculated from the pressure 
at the beginning or end of compression, the corresponding cylin- 
der volumes and the specific volumes corresponding to the pres- 
sure (as obtained from steam- tables). 

The total weight of steam in the cylinder is the weight of steam 
taken into the engine per stroke plus the weight of steam caught 
in the clearance space (cushion steam). The saturation curves 
may now be drawn by plotting the volumes which the total 
weight of steam will occupy at different pressures, assuming it 
to be dry throughout the stroke. 

The quality curve (Fig. 43) shows the condition of the steam 
as the expansion goes on. At any given absolute pressure, the 
volume up to the expansion line shows the volume of the wet 
vapor, while the volume up to the saturation curve shows the 
volume that the weight of the wet vapor would have if it were 
dry. Thus at any given absolute pressure, the ratio of the vol- 
ume of the wet vapor (as given by the expansion line of the 
indicator card) to the total volume of the dry vapor (as obtained 
from the saturation curve) is the measure of the quality of the 
steam. 

Showing the quality by the use of the figure, we have Vol. 



COMBINED INDICATOR CARD OF COMPOUND ENGINE 143 



160 
155 
150 
145 
140 
135 
130 
125 
120 
115 
110 
105 
100 

^95 

cr 
<» 90 

Ft 

D 

f«85 

£ 

§ 80 
o 

S 75 






70 
65 
60 
55 
60 
45 
40 
85 
30 
25 
20 
15 
10 
5 



155 


61* H. 


P. Qua 


lity < 


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Volume- Cu. Ft, 

Fig. 43. — Combined Indicator Diagrams for Compound Engine. 



144 CYCLES OF HEAT ENGINES USING VAPORS 

ab -v- Vol. ac = quality. By laying off this ratio from a hori- 
zontal line to any scale desired as shown, the quality curve may 
be constructed. 

Hirn's Analysis. In the study of steam-engine performance 
the action of the steam in and the influence of the walls of the 
cylinder become a matter of considerable importance. The 
amount of heat lost, restored and transformed into work as a 
result of variation in the condition of the steam throughout the 
engine-cycle can be determined either by means of the entropy- 
temperature diagram, or by calorimetric method based on Hirn's 
theory. 

The calorimetric method, or more popularly called Hirn'r, 
Analysis, was developed by Professor V. Develshauvers-Devy of 
Liege. (See Table of the Properties of Steam by V. Devels- 
hauvers-Devy, Trans. Am. Soc. Mech. Eng., Vol. XI; also Pea- 
body's Thermodynamics.) 

In order to apply Hirn's Analysis to engine testing besides the 
usual readings the following items must be determined : 

i. Absolute pressures at cut-off, release and compression from 
indicator cards, 

2. Per cent of stroke at cut-off, release and compression from 
indicator cards, 

3. Weight of steam per stroke determined by weighing the 
condensed steam or boiler feed water and computing the steam 
used per stroke, 

4. Weight of the cooling water and its temperature at inlet 
and outlet from condenser, 

5. Temperature of the condensed steam. 

Let w = pounds of steam supplied to the cylinder per stroke 

at pressure p and quality x. 

Then the amount of heat brought in by the steam into the 

cylinder per stroke, 

Q = w (h + xL), (191) 

where L = heat of vaporization and h = heat of the liquid 
above the freezing point. 



HIRN'S ANALYSIS 



145 



At the end of compression a certain amount of steam is left 
in the clearance space and this steam mingles with the w pounds 
of steam at admission. Calling the weight of the steam caught 
in the clearance space Wi, the total amount of steam at admission 
is w -f- w\. 

According to thermodynamic laws the addition of heat to a 
substance produces in that substance internal and external 
changes, the internal changes being volume and temperature 
variations which are called intrinsic energy changes, while the 




Volume 
Fig. 44. — The Steam Engine Cycle. 

external changes are changes in external potential energy or work 
done. Thus the steam brought into the cylinder per stroke on 
account of its available heat produces intrinsic energy changes 
and is capable of doing external work in overcoming resistance. 
On the other hand the steam W\ caught in the clearance space at 
compression is able to produce only internal or intrinsic energy 
changes. The intrinsic energy changes can be calculated knowing 
the heat equivalent of internal work or internal latent heat (p) 
and the heat of the liquid Qi) . The external work can be deter- 
mined from the indicator cards. 

Referring to Fig. 44 and calling the heat absorbed by the 
cylinder walls during admission and expansion Q a and Q b ', that 
restored during exhaust and compression Q c and Q d ] the intrinsic 



146 CYCLES OF HEAT ENGINES USING VAPORS 

energy at the points of admission, cut-off, release, and compres- 
sion /1, I 2 , Iz, I *', the external work in heat units AW a , AW b , 
AW C , AWd, during the events of the stroke, where A = t™; 
also if the temperature of the condensed steam is t s and that of 
the cooling water U and U at inlet and outlet, the total weight of 
cooling water used being G, we have the following relations: 

Qa = Q + h ~ h ~ AWa. (192) 

Q b = h-h- AW b . (193) 

Q c = I3 — I4 — wh s — G (h — hi) — AW C . (194) 
Q d = h-h + AW d . (195) 

In equations (192), (193), (194), and (195) Q can be determined 
from Q = w (h + xL). The values of G can be determined by 
weighing cooling water; h , ht, and h s by taking the temperature 
t 0i U, and t s by thermometers and finding h , hi, and h s from steam 
tables. The intrinsic energy at the events of the stroke are calcu- 
lated by means of the following equations: 

h = Wi (h + tfipi), (196) 

h = (wi -J- w) (h 2 + X2P2), (197) 

h = (wi + w) (h z + ^sPs), (198) 

7 4 = Wi (h + ^4p4), (199) 
in which 

W\ — weight of steam caught in clearance space. 

w — weight of steam brought into the cylinder per 
stroke. 
hi, 1h, h z , hi = heat of the liquid at events of stroke. 
Pi, p2, P3, pi = internal latent heats at the events of stroke. 
%i, x 2 , x 3 , x± = quality of steam at events of stroke. 

From the above it is evident that Xi, x 2 , x 3 , #4, and w are un- 
known, these values being determined by the following procedure: 
The volume of w pounds of steam in cubic feet is 

V = w (xu + 0). (200) 

x = quality of steam, u = the increase in volume produced by 
the vaporization of one pound of water of volume a into dry 



CLAYTON'S ANALYSIS OF CYLINDER PERFORMANCE 147 

steam of volume S, or u = S — a. Calling the volume of the 
steam caught in the clearance space Vi and that at cut-off, release 
and compression V 2 , V3, F 4 , 

Vi = Wi(xiUi + a). (201) 

Vi + V 2 = (w + Wi) (x 2 u 2 + a). (202) 

Vi + V 3 = (w + Wi) (X3U3 + a). (203) 

Vi + Va = W\ (X4U4 + <r). (204) 

In the above equations the absolute pressures and volumes can 
be determined from the indicator cards, provided the dimensions 
of the engine and the clearance are known and it is assumed 
# 4 is unity. This assumption can be made without much error, 
as the steam at compression is very nearly dry. If x± is assumed 
equal to 1, then, from equation (204), 

V 1 + F 4 f , 

w l = (205) 

U\ + cr 

The intrinsic energy at the events of the stroke is computed 
after finding x h x 2 , and x 3 from equations (201), (202), and (203) 
by the aid of equation (205).* 

Clayton's Analysis of Cylinder Performance. The indicator 
diagram has been used in determining the economy of the steam 
engine although inaccuracies enter because of cylinder condensa- 
tion. The quantity of steam admitted to the engine per cycle 
can theoretically be determined from the difference between the 
weight of steam accounted for at the point of cut-off and point of 
compression. Little error is introduced in determining the 
weight of steam present in the cylinder at the point of compres- 
sion, for the quality of the steam at that point can be fairly accu- 
rately estimated. To determine the weight of steam present at 
the point of cut-off the quality of the steam must be known. The 
quality of the steam at the point of cut-off is not the same as that 
at admission because of the condensation within the cylinder. 
Thus the study of cylinder performance from the indicator dia- 
gram is made inaccurate. 

* For log form of Hirn's Analysis see Experimental Engineering by Carpenter 
and Diederichs, Seventh Edition, pages 799-806. 



148 CYCLES OF HEAT ENGINES USING VAPORS 

Clayton's analysis was made to determine the relation be- 
tween the quality of the steam at cut-off and other variables. 
From the results of a careful study of the forms of the expansion 
and compression curves which occur in indicator diagrams, it has 
been discovered that the value of n for the expansion curve bears 
a definite relation to the proportion of the total weight of steam 
mixture which was present at the point of cut-off. 

To facilitate the analysis, the indicator diagram is transferred 
to logarithmic coordinates. The equation of the poly tropic 
curve, PV n = C, becomes a straight line when plotted upon 
logarithmic cross-section paper, the value of n becoming the slope 
of the curve. From the new diagram, the slope of the expansion 
and compression curves may be easily determined. The quality 
of the steam at cut-off may then be estimated, knowing the speed 
of the engine, the quality of the steam at admission, the pressure 
of the steam and other values that affect the slope of the curve.* 

PROBLEMS 

1. Assume that 1 lb. of steam of a pressure of 160 lbs. per sq. in. abso- 
lute and a quality of 0.95 performs an ideal Rankine cycle, being exhausted 
at a pressure of 5 lbs. per sq. in. absolute. Compute the quality of the steam 
exhausted, the efficiency of the cycle, and the final volume. 

2. What is the work of an ideal Rankine cycle if the steam initially at 
200 lbs. per sq. in. absolute pressure, superheated 200 F., goes through such 
a cycle with a back pressure of 1 lb. per sq. in. absolute? 

3. One pound of steam at a pressure of 100 lbs. per sq. in. absolute 
with a quality of 0.90 performs an ideal Rankine cycle exhausting at a back 
pressure of 2 lbs. per sq. in. absolute. ' What is the net work and efficiency 
of the cycle? 

4. Two pounds of steam at a pressure of 125 lbs. per sq. in. absolute 
and a volume of 8.34 cu. ft. perform an ideal Rankine cycle. The exhaust 
pressure is 25 lbs. per sq. in. absolute. What is the net work and efficiency 
of the cycle? 

5. One pound of steam at a pressure of 160 lbs. per sq. in. absolute and 
a quality of 0,95 passes through a modified Rankine cycle.- The terminal 

* For details concerning Clayton's "Analysis of the Cylinder Performance 
of Reciprocating Engines," see Bulletin No. 58 of the Engineering Experiment 
Station of the University of Illinois. 



PROBLEMS 149 

pressure is 11 lbs. per sq. in. absolute and the exhaust pressure 5 lbs. per sq. 
in. absolute. What is the efficiency of the cycle? 

6. One pound of steam at a pressure of 200 lbs. per sq. in. absolute and 
200 F. superheat passes through a modified Rankine cycle. The terminal 
pressure is 15 lbs. per sq. in. absolute and the exhaust pressure 10 lbs. per 
sq. in. absolute. What is the net work and efficiency of the cycle? 

7. Two pounds of steam at a pressure of 125 lbs. per sq. in. absolute and 
volume of 8.34 cu. ft. pass through a modified Rankine cycle. The terminal 
pressure is 20 lbs. per sq. in. absolute and the exhaust pressure 15 lbs. per 
sq. in. absolute. What is the efficiency, net work, and heat added to the 
cycle? 

8. A non-condensing steam engine receives steam of 0.95 quality and 
100 lbs. per sq. in. absolute pressure. The exhaust pressure is 15 lbs. per 
sq. in. absolute and the steam consumption of the engine is 30 lbs. per 
indicated horse power per hour. What is the thermal efficiency of the 
engine? What is the efficiency of the ideal Rankine cycle operating under 
the same conditions? What is the efficiency of the Carnot cycle operating 
between the same temperature limits? 




Throat or smallest 
section of nozzle 



CHAPTER IX 

FLOW OF FLUIDS 

Flow through a Nozzle or Orifice. Thermodynamic prob- 
lems embrace the measurement of the flow of air or of a mix- 
ture of a liquid and vapor through a nozzle or orifice. In the 

nozzle shown in Fig. 45, let A and B 
be two sections through which the sub- 
stance passes. At A let a pressure of 
Pi be maintained and at B a pressure 
of P 2 . To maintain the constant pres- 
sure at A of Pi let more substance be 
Fig. 45. -- - Typical Nozzle added, while at B allow enough of the 
for Expanding Gases and Va- substance to be discharged so that the 
pors * constant pressure of P 2 is maintained. 

The quantity of energy and the mass passing into the section A 
must be accounted for at section B and the relation of these 
quantities will determine the change of the velocity of the sub- 
stance. 

After uniform conditions have been established in the nozzle, 
the same mass entering at A must be discharged at B during 
the same time. Thus any mass may be considered as a work- 
ing basis, but as a rule one pound of the substance is used. All 
formulas refer to one pound, unless another mass is definitely 
stated. 

The same quantity of energy discharged at B must enter at 
A unless heat is added to or taken from the substance between 
the sections A and B. Thus the general formula is derived: 

Energy carried by substance at B = energy carried by sub- 
stance at A + energy added between the sections A and B. 

The energy carried by the substance at the entering or dis- 
charge end of the nozzle is made up of three quantities: (1) the 

150 



FLOW THROUGH A NOZZLE OR ORIFICE 151 

amount of work necessary to maintain constant pressure at each 
end of the nozzle; (2) the internal energy of the substance, 
(3) the kinetic energy stored in the substance because of the 
velocity which it has when passing the section. 
The amount of work necessary to maintain a constant pres- 

sure (pounds per square foot) of Pi at A or of P 2 at B is — ^ or 

778 
p „. 

—~, where i\ and % are the volumes of one pound of the sub- 
778 

stance at A and B respectively. 

The internal or " intrinsic " energy in B.t.u. per pound (I H ) 

of the substance passing A or B, calculating from 32 F. is for: 

Air or similar gases, \- ? (206) 

5 ' 778X040 

Liquid, h t (207) 

/ 72 \ 

Liquid and Vapor, h + x ( L )> (208) 

H V 2 g X 778/ 

/ V 2 \ 

Superheated Vapor, h + \L ) 

F V 2 g X 778/ 

+c y (r SUP -r sa t.)- jP(w 7 z;sat) , (209) 

778 

V 2 

where is the kinetic energy in B.t.u. per pound of the 

2 g X 778 

substance as it passes a section, V = velocity in feet per second 

and g = 32.2 feet. 

If Q represents in B.t.u. per pound the heat units added to the 

substance between the sections A and B (Fig. 45), the energy 

equation for air and similar gases can be found by equating the 

total heat energy put in at A plus the energy added between 

A and B to the energy discharged at B. This general formula 

reduces to 

F 2 2 - TV = 2 g [j^ (P 1 v 1 - P 2 v 2 ) + 778 <2~j 
or = 2 g X 778 [C 9 (Xi - T 2 ) + Q]. (210) 

* See equation (61). 



152 FLOW OF FLUIDS 

This thermodynamic equation is the usual form for the flow 
of air or similar gases. 

The energy equation for superheated steam can be derived 
by the use of the same general formula stated above, on the 
assumption that the substance is superheated at A and B. This 
formula reduces to 

-= [h + L\ + C p (r si ,p. — Taa,t)i] 

2 g X 778 

— [h 2 + l 2 + c P (r SUP . — r sa t.) 2 ] + Q, (211) 

or V 2 2 — V1 2 = 2 g X 778 [Tabular heat content* — Tabu- 

lar heat content^ + Q], (212) 

From a thermodynamic standpoint, the relation between the 
initial and final condition is that of adiabatic expansion when all 
the heat which disappears as such is used in changing the veloc- 
ity, provided the nozzle is properly shaped and Q is zero. The 
diagram in Fig. 46 represents this condition of affairs on a tem- 
perature-entropy diagram for air and similar substances. The 
area acdf is C v (T 2 — 492), and the area abef is C p ( Ji — 492). 
The quantity of heat energy changed into kinetic energy is there- 
fore the area bcde and is the difference between the internal 
energy in the substance at the beginning and end of the opera- 
tion, together with the excess of work done to maintain the pres- 
sure of P± at A over the pressure of P 2 at B. The line cd would 
incline to the right if heat were added in the nozzle, since the 
effect would be to increase the velocity or increase the area cdeb. 
The line cd would incline to the left from c if heat were lost, 
since the area cdeb would decrease. 

In Fig. 47 the diagram represents the conditions for super- 
heated vapor. The area aa'cdf represents the heat required to 
raise the substance from a liquid at 32 F. to superheated vapor 
at the temperature of Tisup. The area ab'bdf is the heat con- 
tent at b, the final condition. The area a'cbb' represents, there- 
fore, the heat available for increasing the velocity. The areas 

* Tabular heat content means the total heat of superheated steam as read 
from tables of the properties of superheated steam. 



FLOW THROUGH A NOZZLE OR ORIFICE 



153 



representing the heat available for increasing the velocity in Fig. 

46 and Fig. 47 are shown by the cross-hatched area in Fig. 48 

and are really the representation of the work done (theoretically) 

in an engine giving such an indicator diagram. 

Evidently the greater the drop 
in pressure, the greater will be the 
cross-hatched area in Fig. 48 and 
the greater will be the velocity, 



T 2 




c n 


T, 




g\ 




V 




a/ 






492 


T.- Ent.- Diagram 


i 

1 


/ 


la 


d\\ 



t^up. OJ 
Tjsat. a' 1 




T 2 sup. / \ 


6 


T2 sat. / \ 


r 

U T.-Ent.-Diagram 


d 



Fig. 46. — Temperature- 
entropy Diagram of Heat 
Available in Air. 



Fig. 47. — Temperature-entropy Dia- 
gram of the Heat Available in Superheated 
Steam for Increasing Velocity. 



regardless of the substance. The line ab in Fig. 49 represents 

p 
the velocity curve with V as ordinates and -^ as abscissas, but 

" 1 

since with any substance ex- ! i 
panding the weight of a cubic 
foot decreases as the pressure 
drops, the line cd will repre- 
sent to some scale not here, 
determined the weight of a 
cubic foot at any discharge 
pressure 




Fig. 48. — Heat (Work) Available for 
Increasing Velocity. 



Since the product of the area through which the discharge 
takes place, the velocity and the weight per cubic foot of the 



154 



FLOW OF FLUIDS 



substance is equivalent to the weight in pounds of the substance 
discharged per second, the product of the ordinates at any point 
of the curves ab and cd is proportional to the weight discharged 
from a pressure of P\ to a condition where the pressure is P 2 . 
The line ceb represents this product. Evidently there is some 
low pressure into which the weight discharged per square foot 
will be a maximum, and this will be the pressure corresponding 
to the high point e on the curve. 



a 


















V 














e 




















\f 


v in.p 


^ 
















/4 




































d 










\ c 


& 








































^ 




















f 


$? 






































l/c 




















b 



.2 



.5 



.6 .7 



.9 1.0 



Fig. 49. — Illustrative Curves of Weight, Discharge and Velocity. 

Weight per Cubic Foot. From the formula for adiabatic ex- 
pansion the weight per cubic foot can be obtained if the sub- 
stance is similar to air.* The general formula (applied to air) is 

(213) 



(214) 



* The exponent in the formula is the ratio of the specific heats (of air in this 
case). 



P^l-40 B 


P2^- 40 , 


q be reduced to 






1 
1 P 2 14 X Pi, t 

Pi 1 - 4 X RTi 


or 


-,- p 0.286 p 0.714 

v 2 RTi 



FLOW OF AIR THROUGH AN ORIFICE 155 

which is the weight in pounds per cubic foot of discharge. Pres- 
sures are in pounds per square foot. If the supply to the nozzle 
is from a large reservoir so that V± can be taken as zero then 
the discharge velocity is 



T 0.4 

v, - V/Vx^Wx-g) 



v, = 109.6 \/ r, [1 - (0' 286 ]- (21s) 

All quantities on the right-hand side of this equation must be 
obtained from the data of tests. • Weight in pounds discharged 
through the area A (in square feet) is 



P 0.286 P„0-714 

w = A x Fl Jj xi 

Kl 1 



o 9 .6v/ r,[i - (0""} (216) 



Maximum Discharge. This weight is a maximum when 

— - = o or when P 2 = 0.525 Pi. The maximum quantity of air 
aP 2 

will be discharged when the low pressure is 52.5 per cent of the 
high pressure. 

Shape of Nozzle. See page 167, on the flow of steam. 

Flow of Air through an Orifice. Air under comparatively high 
pressure is usually measured in practice by means of pressure 
and temperature observations made on the two sides of a sharp- 
edged orifice in a diaphragm. The method requires the use of 
two pressure gages on opposite sides of the orifice and a ther- 
mometer for obtaining the temperature ti at the initial or higher 
pressure pi. The flow of air w, in pounds per second, may then 
be calculated by Fliegner's formulas : 

w = 0.530 X f X a— t^z when pi is greater than 2 p 2 , (217) 

vTi 

w = 1.060 Xf X a 1 /P2 (Pi ~ P2) w hen p x is less than 2 p 2 , (218) 

V Ti 



156 FLOW OF FLUIDS 

where a is the area of the orifice in square inches, f is a coeffi- 
cient, Ti is the absolute initial temperature in degrees Fahren- 
heit at the absolute pressure pi in the " reservoir or high-pressure 
side" and p 2 is the absolute discharge pressure, both in pounds 
per square inch. When the discharge from the orifice is directly 
into the atmosphere, p2 is obviously the barometric pressure. 

Westcott's and Weisbach's experiments show that the values 
of f are about 0.925 for equation (217) and about 0.63 for equa- 
tion (218). 

For small pressures it is often desirable to substitute manom- 
eters for pressure gages. One leg of a U-tube manometer can be 
connected to the high-pressure side of the orifice and the other leg 
to the low-pressure side. Valves or cocks are sometimes inserted 
between the manometer and the pipe in which the pressure is to 
be observed for the purpose of " dampening" oscillations. This 
practice is not to be recommended as there is always the possi- 
bility that the pressure is being throttled.* A better method is 
to use a U-tube made with a restricted area at the bend between 
the two legs. This will reduce oscillations and not affect the 
accuracy of the observations. 

Discharge from compressors and the air supply for gas engines 
are frequently obtained by orifice methods. 

When pi — p 2 is small compared with pi, the simple law of 
discharge f of fluids can be used as follows: 

* Report of Power Test Committee, Journal A. S.M.E., Nov., 191 2, page 1695. 
f If the density is fairly constant, 

144 pi v£ 144 p 2 Vq2 
s + 2g s i "2g' 

where Vi is the velocity in feet per second in the "approach" to the orifice, and v is 
the velocity in the orifice itself. Since Vi should be very small compared with v , 

Vq2 144 (p x - pa) 
2g s ' 



Vo = V /: 



w 



2 g X 144 (p x - p 2 ) . 
= favosa fas y ^^fo-p.) , 



FLOW OF AIR THROUGH AN ORIFICE 157 

fa 



w = V 2 g X 144 (Pi - P2) s, (219) 

144 

where f is a coefficient from experiments, g is the acceleration 
due to gravity (32.2), and s is the unit weight of the gas meas- 
ured, in pounds per cubic foot, for the average of the initial and 
final conditions of temperature and pressure. If the difference 
in pressure is measured in inches of water h with a manometer, 
then 

144 (pi — p 2 ) = — — X h (expressed in terms of pounds per square 

foot), 



w = y 2 ghs X — — (pounds per second), (220) 

144 ^2 

where 62.4 is the weight of a cubic foot of water (density) at 
usual "room" temperatures. 

This equation can also be transformed so that a table of the 
weight of air is not needed, since by elementary thermodynamics 
144 pv = 53.3 T, where v is the volume in cubic feet of one 
pound and T is the absolute temperature in Fahrenheit. Since 
v is the reciprocal of s, then 

s = 144 p + 53.3 T, 




and w = 0.209 fa V/^ ( 221 ) 



Here p and T should be the values obtained by averaging the 
initial and final pressures and temperatures. Great care should 
be exercised in obtaining correct temperatures. For accurate 
work, corrections of s for humidity must be made.* 

For measurements made with orifices with a well-rounded 
entrance and a smooth bore so that there is practically no Con- 



or w = fa VsTg x 144 (pi — p 2 )s. 

Professor A. H. Westcott has computed from accurate experiments that the 
value of the coefficient f in these equations is approximately 0.60. 

* Tables of the weight of air are given on page 181 and tables of humidity on 
page 368 in Moyer's Power Plant Testing (2d Edition). 



158 FLOW OF FLUIDS 

traction of the jet the coefficient f in equations (217) and (218) 
may be taken as 0.98. In the rounding portion of the entrance 
to such a nozzle the largest diameter must be at least twice the 
diameter of the smallest section. For circular orifices with 
sharp corners Professor Dalby * stated that the coefficient for 
his sharp-edged orifices in a thin plate of various sizes from 1 inch 
to 5 inches in diameter was in all cases approximately 0.60; 
and these data agree very well with those published by Durley.j 

When p 2 -J- pi = 0.99 the values obtained with this coefficient 
are in error less than J per cent; and when p 2 -f- pi = 0.93 the 
error is less than 2 per cent. 

Receiver Method of Measuring Air. None of the preceding 
methods are adaptable for measuring the volume of air at high 
pressures as in the case of measuring the discharge in tests of 
air compressors. Pumping air into a suitably strong receiver 
is a method often used. The compressor is made to pump 
against any desired pressure which is kept constant by a regu- 
lating valve on the discharge pipe: 

Pi and P 2 = absolute initial and final pressures for the test, 
pounds per square inch. 

Ti and T 2 = mean absolute initial and final temperatures, de- 
grees Fahrenheit. 

Wi and w 2 = initial and final weight of air in the receiver, 
pounds. 

V = volume of receiver, cubic feet. 

PiV = wRTi, and P 2 V = w 2 RT 2 , 'where R is the constant 
53.3 for air, then weight of air pumped 



L (h _ h\ 
i.3 VT, Tj 



w 2 - wi = — - (—- — )• (222) 

In accurate laboratory tests the humidity of the air entering 
the compressor should be measured in order to reduce this 

* Engineering (London), Sept. 9, 19 10, page 380, and Ashcroft in Proc. Insti- 
tution of Civil Engineers, vol. 173, page 289. 

f Transactions American Society of Mechanical Engineers, vol. 27 (1905), 
page 193. 



FLOW OF VAPORS 



159 



weight of air to the corresponding equivalent volume at atmos- 
pheric pressure and temperature. 

The principal error in this method is due to difficulty in measur- 
ing the average temperature in the receiver. Whenever practi- 
cable the final pressure should be maintained in the receiver at 
the end of the test until the final temperature is fairly constant. 

Flow of Vapors. In Fig. 45 suppose the sections A and B 
are so proportioned that the velocity of the substance passing 
section A is the same as that at section B. Such a condition 
might arise in a calorimeter or in the expansion of ammonia 
through a throttling or expansion 
valve, as in an ice machine. The 
pressure at A will be Pi which is 
greater than the pressure at B of P 2 . 
Fig. 50 represents the entropy dia- 
gram for such a condition. As the 
pressure falls from Pi to P 2 the maxi- 
mum heat available to produce veloc- 
ity through the nozzle is the area 
acde. The value of the quality, rep- 
resented bv the symbol x for the 
substance after leaving the nozzle FlG ' S °'~ Q ^%^ T n ° Ve ~ 
corresponds to that of point c and 

the area acde is the excess kinetic energy represented by the 
increased velocity. This excess kinetic energy is destroyed by 
coming into contact with the more slowly moving particles at 
B and with the sides of the vessel. The area acde is equal to 
(h a + x a L a ) — (h c + x c L c ) and the relation of x a to x c is obvi- 
ously adiabatic. The area ohdbf equals area oheag (thus the 
heat content at b is the same as at a). The location of b can be 
found as follows: 





I 6 ' Pi 


a 






Id 




c b 




/ *" 






h 


T-Ent.-Diagram 











9 


f 



Xb 



Ha ~T~ Xa-L*a »^& 



(223) 



The curve shown in Fig. 51 represents the discharge of a mix- 
ture of steam and water (x = 0.6 at 100 pounds per square inch 
absolute pressure) into a vessel having the pressures shown. The 



i6o 



FLOW OF FLUIDS 



points on this curve cannot be determined by entropy tables. 
At ioo pounds per square inch pressure the total heat of the wet 
steam is h + 0.6 L or 298.5 + 0.6 X 887.6 = 831. 1 B.t.u. per 
pound. 



2.00 



1.75 



1.50 



o 

a 

o 

o 

£ 1.25 

o 

o 1 

is 

^ 1 1.00 

ra 03 

a .75 






.50 



.25 

















4/ 












to/ 

7 







































































10 20 30 40 50 60 70 80 90 100 
Discharge Pressureln. lbs. per sq. in. 

Fig. 51. — Discharge of Steam Under Various Pressures. 

At 60 pounds per square inch absolute pressure the total heat 
may be found by the use of the entropy diagram shown in 
Fig. 52. The entropy values are taken directly from steam 
tables. The entropy for the initial point is, then, 
ab = 0.4748 + 0.6 X 1. 1273 = 1.1512. 

The distance 

dc = ab — 0.4279 = 0.7233, 

and x at 60 pounds is : 

dc 0.7233 

= = 0.595. 

1. 2154 1. 2154 

The total heat at 60 pounds pressure is 

h + 0.595 L = 262 -4 + 0.595 X 914.3 = 805.4. 



FRICTION LOSS IN A NOZZLE 



161 



The velocity of flow is 



V2 x 32.2 x 778 (831. 1 — 805.4) = 1135 feet per second. 

The volume of one pound is 
0.016 (1.0 — 0.595) + 7.166 X 0.595 = 4.27 cubic feet, 
and the weight per cubic foot is 



4.27 



= 0.2342 pound. 




1.1273- 



1.1512- 



1.2154- 



T-Ent.-Diagram 




The weight discharged per square 

inch per second is 

1135 X 0.2342 _ , 

— — ^- = 1. 8 s pounds. 

144 F 

Velocity of Flow as Affected by 
Radiation. Fig. 53 shows the radia- ,..,.,....., 

tion losses. The condition at entrance tropy Diagram for Calculating 

is represented at a and the area acde the Weight of Discharge of 

represents the quantity of heat lost team * 

by radiation. Area aefg represents the velocity change while 

the point e represents, the condition of the moving substance. 

If, after passing through the nozzle, 
the velocity is reduced to that of en- 
trance, a point located as at b will 
represent the condition of the sub- 
stance. This point would be so located 

that 

area aefg . , x 

(224) 




T.- Ent.- Diagram 



eb = 



L f 



Friction Loss in a Nozzle. Fig. 54 

shows the friction loss. The energy 
converted into heat by friction varies 
with the square of the velocity. In 
Fig. 53. — Diagram illustrating this figure, a is the initial condition 

Radiation Loss in Nozzle. and acfg ^ ^ energy available for 

change in velocity provided there is no frictional loss. The ratio 
of the areas acde to acfg is the proportional loss by friction. 
The point c represents the condition of the substance at the 



l62 



FLOW OF FLUIDS 



return of the friction heat to the substance. The heat is 
returned in exactly the same way as if it came from an outside 

source. The distance ch is the area 
acde -T- L c . The area edfg represents 
the energy expended in the velocity 
change and the point h represents the 
state of the substance at the point of 
discharge. 

This condition is found existing in 
the fixed nozzle of most turbines. 
Point a represents the condition on 
the high-pressure side of the nozzle 
and point h, the low-pressure side. 
The absolute velocity of discharge is 



T.- Ent.- Diagram 



n I d 



a 



Fig. 54. — Diagram Illustrating really caused by the energy repre- 
Friction Loss in Nozzle. S ented by the area edfg. 

Impulse Nozzles. Suppose that the substance is discharged 
with an absolute velocity corresponding to the area edfg (Fig. 
54), and that it passes into a mov- 
ing nozzle, having the same pressure 
on the discharge as on the intake 
side. The energy represented by 
the area edfg would be used up in 
the following ways: (1) by the fric- 
tion in moving nozzle; (2) residual 
absolute velocity; (3) and energy 
used in driving the moving nozzle 
against the resistance. 

Fig. 55 shows the quantities used 
up by each. Point a represents the 
condition of the substance before 
passing into the fixed nozzle while 
point h shows its condition leaving 
the fixed nozzle, the velocity corre- 
sponding to the area edfg. Area klde represents the energy 
used up in friction in the moving nozzle; area klnm residual 




T.- Ent.- Diagram 



Fig. 55. — Diagram of Heat Losses 
in a Steam Nozzle (Impulse). 



TURBINE LOSSES 



163 



velocity after leaving moving nozzle and area mnfg represents 
useful work used in moving the nozzle against its resistance. 
The condition of the substance leaving the nozzle is shown at 
q and not at h, the distance h r- q being the area edlk divided 
by Ln. The substance leaves the moving nozzle with a velocity 
corresponding to the area klnm and it will have done work 
corresponding to the area mnfg. 

Turbine Losses. Fig. 56 is a simple velocity diagram show- 
ing, for an impulse nozzle such as occurs in many turbines, the 
relative value of those various losses. A is a stationary nozzle 




e v 



Fig. 56. — Impulse Nozzle and Velocity Diagrams. 

discharging against the movable blades B. The path of the steam 
is shown by the dotted line. The line db marked v represents 
the velocity of discharge of the stationary nozzle, which makes 
an^angle a with the direction of motion of the moving blades. 
Call ev the velocity of the moving blades, then h is the amount 
and direction of the relative velocity of the steam over the 
surface of the moving blades. It loses a portion of this velocity 
as it passes over the surface of the blades and Ih becomes the 
actual relative velocity of discharge. The direction of Ih is de- 
termined by the discharge edge of the moving blades, the angles 
a and /3 being as shown. The residual absolute velocity is rep- 
resented by r. 
The total energy equivalent of the velocity developed 



in B.t.u. per pound = 



2g 



(225) 



164 



FLOW OF FLUIDS 



w 

The residual energy per pound = — 

2 i 



(226) 



n 



T.- Ent.- Diagram 



Reaction Nozzles. When the substance leaving the station- 
ary nozzle passes into a moving nozzle having the pressure at 
the intake greater than at the discharge, the conditions differ 
from those just discussed. The velocity in this case is changed 
in passing through the moving nozzle. In the equations given 
in the previous discussion it was assumed that the moving nozzles 

were entirely filled with the sub- 
stance, and when partly filled in the 
expanding portion, coefficients of 
correction were applied, but in this 
case the nozzles should be so de- 
signed that the substance entirely 
fills them, as the corrections are un- 
known. 

In Fig. 57 the lines of Fig. 55 are 
reproduced together with those re- 
lating directly to the reaction nozzle. 
Point a corresponds to the condition 
on entering the stationary nozzle, 

Fig. 57. -Diagram of Heat Losses •<. h ^ condition on leaving ft 
in a Steam Nozzle (Reaction). . 

with a velocity corresponding to the 

area edfg. Point k represents the pressure at the discharge end 

of the moving nozzle, and if no friction losses or impact loss occur 

in the moving nozzle, point k would represent the condition of the 

discharged substance and the area egmkhd would be accounted 

for as useful work done and residual velocity. But since friction 

losses and impact loss do occur a portion of this area edhkno can 

be set aside to represent these losses, a portion noqp represents 

the residual velocity, while the remaining area pqgm represents 

the useful work done. 

The condition of the substance leaving the moving nozzle is 

, , ,, ,. . , , 1 • area edhkno . 
given by k, the distance kl being Area onpq rep- 




Li 



resents the residual velocity. 



INJECTORS 165 

Coefficient of Flow. Few experiments have been carried on for 
determining the flow of steam in nozzles proportioned for maxi- 
mum discharge. For a nozzle having a well rounded entrance 
and with the parallel portion of least diameter from 0.25 to 1.5 
times the length of the converging entrance, the coefficient of 
discharge is about 1.05. For properly shaped entrances and for 
areas of orifices between 0.125 square inch and 0.75 square inch 
the coefficient of discharge varies from 0.94, the two pressures 
being nearly alike, to unity, the ratio of the pressures being 0.57. 
For an orifice through a thin plate the coefficient is about 0.82, 
the ratio of the pressures being 0.57. 

Injectors. In an injector, steam enters at A in Fig. 58 at 
the pressure of the supply. The quantity of water entering at 
C, the cross-section of the pipe, and the pressure of the water 
determine the pressure at B. At D 
the pressure should be zero (atmos- 
pheric) or equal to the pressure in 

the water supply pipe to which D Fig. 58. --Essential Parts of an 

may be connected. 'The total hy- 
draulic head should exist as velocity head at this point. At E 
the pressure should be sufficient to raise the check valve into 
the boiler and the velocity sufficient to carry the intended 
supply into the discharge pipe. 

The shape of the nozzle from A to B should be such as to con- 
vert the energy in the steam at A into velocity at B. At B the 
water and steam meet, condensing the steam, heating the water 
and giving to the water a velocity sufficient to carry it through 
the nozzle B-D. 

All the energy accounted for at A and C must be accounted for 
at E. The heat lost by radiation may be neglected. The veloc- 
ity at any section of the nozzle equals 

volume passing in cubic feet 
area of section in square feet 

or V A = — » where a = area at any section corresponding to 
velocity V and a a = area at section A . 




1 66 FLOW OF FLUIDS 

Weight of Feed Water Supplied by an Injector per Pound of 
Steam. Assuming the steam supply to be dry the heat units 
contained in the steam and feed water per pound and the heat 
in the mixture of steam and feed water per pound may be easily 
calculated. 

Knowing the rise of temperature of the water passing through 
the injector and neglecting radiation losses, the pounds of feed 
water supplied per pound of steam used by the injector may be 
obtained. Thus : 

Heat units lost by steam = Kinetic energy of jet + Heat 
units gained by feed water. 

The kinetic energy of jet may be neglected since it is very 
small, then, 

H — h/ = w (km — hf) 

H -h f , v 

W = =-> (227) 

rim — hf 

where w = the weight of feed water lifted per pound of steam. 
h m = heat of liquid of mixture of condensed steam and 

feed water. 
hf = heat of liquid of entering feed water. 

Thermal Efficiency of Injector. The thermal efficiency of an 
injector neglecting radiation losses is unity. All the heat ex- 
pended is restored either as work done or in heat returned to the 
boiler. 

Mechanical Efficiency of Injector. The mechanical work per- 
formed by the injector consists in lifting the weight of feed water 
and delivering it into the boiler against the internal pressure. 
The efficiency, considering the injector as a pump, is 

Work done 
B.t.u. given up by steam to perform the work 

or 

where U = [wl s + (w + 1) l P ] -f- 778 (in heat units). 



FLOW OF STEAM THROUGH NOZZLES 



167 



Is = 

W = 



pressure head corresponding to boiler gage pressure, 

in feet, 
suction head in feet, 
pounds of water delivered per pound of steam. 



Orifice Measurements of the flow of steam are sometimes used 
for ascertaining the steam consumption of the "auxiliaries" in a 
power plant. This method commends itself particularly because 
of its simplicity and accuracy. It is best applied by inserting a 
plate § inch thick with an orifice one inch in diameter, with square 
edges, at its center, between the two halves of a pair of flanges 
on the pipe through which the steam passes. Accurately cali- 
brated steam gages are required on each side of the orifice to 
determine the loss of pressure. The weight of steam for the 
various differences of pressure may be determined by arranging 
the apparatus so that the steam passing through the orifice will 
be discharged into a tank of water placed on a platform scales. 
The flow through this orifice in pounds of dry saturated steam 
per hour when the discharge pressure at the orifice is 100 pounds 
by the gage is given by the following table: 



Pressure drop, 
lbs. per sq. in. 


Flow of dry steam 
per hour, lbs. 


Pressure drop, 
lbs. per sq. in. 


Flow of dry steam 
per hour, lbs. 


1 
2 

I 
2 

3 

4 


430 

6i5 

930 

1200 

1400 


5 
10 

15 
20 


1560 
2180 
2640 
3050 







Flow of Steam through Nozzles. The weight of steam dis- 
charged through any well-designed nozzle with a rounded inlet, 
similar to those illustrated in Figs. 59 and 60, depends only on 
the initial absolute pressure (Pi), if the pressure against which 
the nozzle discharges (P 2 ) does not exceed 0.58 of the initial 
pressure. This important statement is well illustrated by the 
following example. If steam at an initial pressure (Pi) of 100 
pounds per square inch absolute is discharged from a nozzle, the 



i68 



FLOW OF FLUIDS 




weight of steam flowing in a given time is practically the same 
for all values of the pressure against which the steam is dis- 
charged (P 2 ), which are equal to or less than 58 pounds per square 
inch absolute. 

If, however, the final pressure is 
more than 0.58 of the initial, the weight 
of steam discharged will be less, nearly 
in proportion as the difference between 
the initial and final pressures is re- 
duced. 

The most satisfactory and accurate 
formula for the " constant flow " con- 
dition, meaning when the final pressure 
is 0.58 of the initial pressure or less, is 
the following, due to Grashof,* where w 
is the flow of steam f (initially dry 
saturated) in pounds per second, A Q is the area of the smallest 
section of the nozzle in square inches, and Pi is the initial abso- 
lute pressure of the steam in pounds per square inch, 




Fig. 59. — Example of a Well- 
designed Nozzle. 



or, in terms of the area, 



w = 



A, = 



ApPj 
60 



60 w 



97 



Pi 



97 



(229) 



(230) 



* Grashof, Theoretische Maschinenlehre, vol. 1, iii; Hiitte Taschenbuch, vol. 1, 
page 2>2>Z- Grashof states the formula, 

w = 0.01654 A oPr 9696 , 

but the formula given in equation (229) is accurate enough for all practical uses. 

t Napier's formula is very commonly used and is accurate enough for most 
calculations. It is usually stated in the form 



w = 



70 



where w, Pi, and Ao have the same significance as in Grashof s formula. The 
following formula is given by Rateau, but is too complicated for convenient use: 

w = 0.001 A0P1 [15.26 — 0.96 (log Pi + log 0.0703)]. 
Common or base 10 logarithms are to be used in this formula. 



FLOW OF STEAM THROUGH NOZZLES 



169 



These formulas are for the flow of steam initially dry and 
saturated. An illustration of their applications is given by the 
following practical example. 

Example. The area of the smallest section (A ) of a suitably 
designed nozzle is 0.54 square inch. What is the weight of the 







DeLaval Type. 




Nozzle* 

Diaphragm, 



Curtis Type. 

Fig. 60. — Examples of Standard Designs of Nozzles. 

flow (w) of dry saturated steam per second from this nozzle when 
the initial pressure (Pi) is 135 pounds per square inch absolute 
and the discharge pressure (P 2 ) is 15 pounds per square inch ab- 
solute? 
Here P 2 is less than 0.58 Pi and Grashof's formula is applicable, 



170 FLOW OF FLUIDS 

O.54 (l3$) -97 
or, w = D ^ ) OJ/ > 

60 

0.54 X 1 16.5* , , 

w = __^z — ^l_ = L040 pounds per second. 

60 

When steam passes through a series of nozzles one after the 
other as is the case in many types of turbines, the pressure 
is reduced and the steam is condensed in each nozzle so that 
it becomes more moist each time. In the low-pressure nozzles 
of a turbine, therefore, the steam may be very wet although 
initially it was dry. Turbines are also sometimes designed to 
operate with steam which is initially wet, and this is usually the 
case when low-pressure steam turbines are operated with the 
exhaust from non-condensing reciprocating engines. In all these 
cases the nozzle area must be corrected for the moisture in the 
steam. For a given nozzle the weight discharged is greater 
for wet than for dry steam; but the percentage increase in the 
discharge is not nearly in proportion to the percentage of mois- 
ture as is often stated. The general equation for the theoretic 
discharge (w) from a nozzle" is in the form f 

* The flow (w) calculated by Napier's formula for this example is w = -^ — 

70 

= 1. 04 1 pounds per second. 

f The general equation for the theoretic flow is 



/ r 2 g+1 -1 



where the symbols w, A , Pi, and g are used as in equations (228) and (229). P 2 
is the pressure at any section of the nozzle, vi is the volume of a pound of steam at 
the pressure Pi, and k is a constant. The flow, w, has its maximum value when 



2 i-fl 

I 



m-m 

is a maximum. Differentiating and equating the first differential to zero gives 

1 

il = ( 2 V" 1 

Pi U + i/ ' 

P 2 is now the pressure at the smallest section, and writing for .clearness P for 
P 2 , and substituting this last equation in the formula for flow (w) above, we have 



FLOW OF STEAM 1 71 

fp 

w = K K /—, (231) 

where Pi is the initial absolute pressure and Vi is the specific vol- 
ume (cubic feet in a pound of steam at the pressure Pi). Now, 
neglecting the volume of the water in wet steam, which is a usual 
approximation, the volume of a pound of steam is proportional 
to the quality (xi). For wet steam the equation above becomes 
then 






(232) 



The equation shows, therefore, that the flow of wet steam is 
inversely proportional to the square root of the quality (xi). 
Grashof's equations can be stated then more generally as 

w = ^r_ <*»> 

. 60 W A / V 

= _ p^ — (234) 

These equations become the same as (229) and (230) for the 
case where X\ = 1. 

Flow of Steam when the Final Pressure is more than 0.58 of 
the Initial Pressure. For this case the discharge depends upon 
the final pressure as well as upon the initial. No satisfactory 
formula can be given in simple terms, and the flow is most easily 
calculated with the aid of the curve in Fig. 61, due to Rateau. 



-Wm?M 



Now regardless of what the final pressure may be, the pressure (P ) at the smallest 
section of a nozzle (Aq) is always nearly 0.58 Pi for dry saturated steam. Making 
then in the last equation P = 0.58 Pi and putting for k Zeuner's value of 1.135 
for dry saturated steam, we may write in general terms the form stated above, 



V Vi 



where K is another constant. See Zeuner's Theorie der Turbinen, page 268 (Ed. of 
1899)- 



172 



FLOW OF FLUIDS 



This curve is used by determining first the ratio of the final to 

p 
the initial pressure — -, and reading from -the curve the corre- 

sponding coefficient showing the ratio of the required discharge 
to that calculated for the given conditions by either of the equa- 
tions (229) or (233). The coefficient from the curve times the 



1.0 










































1 

1 








— ' 












































_ qT 












,9 










































lC 




















































I- 












.8 

S§' 7 


























































































































































































































1 1- 6 














































































































* * 

ja <3 .5 
-? 

+3 CO 

§ £ A 
^ 2 

& a 

<3 n 

© r° .3 

O PH *° 






































































































































































































































































































































.2 














































































































i 














































































































:o 

























































uo 



.8 .7 «£ 

Batio of FfnaT to Initial PressureJ^L , 

Pi 



.5 



Fig. 61. — Coefficients of the Discharge of Steam when the Final Pressure is 
Greater than 0.58 of the Initial Pressure. 

flow calculated from equations (229) or (233) is the required re- 
sult. Obviously the discharge for this condition is always less 
than the discharge when the final pressure is equal to or less 
than 0.58 of the initial. 

Length for Nozzles. The length of the nozzle is usually made 
to depend only on the initial pressure. In other words, the 
length of a nozzle for 150 pounds per square inch initial pressure 
is usually made the same for a given type regardless of the final 
pressure. And if it happens that there is crowding for space, 
one or more of the nozzles may be made a little shorter than 
the others. 



UNDER- AND OVER-EXPANSION 



m 



Designers of De Laval nozzles follow practically the same 
" elastic " method. The divergence of the walls of non-con- 
densing nozzles is about 3 degrees from the axis of the nozzle, 
and condensing nozzles for high vacuums may have a divergence 
of as much as 6 degrees * for the normal rated pressures of the 
turbine. 

One of the authors has used successfully the following empirical 
formula to determine a suitable length, L, of the nozzle between 
the throat and the mouth (in inches) : 



L = V I5 A Q j 



(235) 



where A is the area at the throat in square inches. 

Under- and Over-expansion. The best efficiency of a nozzle is 
obtained when the expansion required is that for which the nozzle 



ao 
$0 

P « ° 
'» & 

o * 

« I 
So 

.2 « 
1*2 



* 





















— 








r+- 
























/ 






















































/ 






























1 






















































1 






















































1 
i 






















































1 












































































































i 






















































1 












































































































-4- 





























30 25 20 15 10 ^ 5 

"Percentage Nozzle is too Small^ 

at Mouth (Under Expansion) 



5 10 15 20 25 30 

Percentage Nozzle is too Larger 

at Month (Over Expansion) 



Fig. 62. — Curve of Nozzle Velocity Loss. 



was designed, or when the expansion ratio for the condition of 
the steam corresponds with the ratio of the areas of the mouth 
and throat of the nozzle. A little under-expansion is far better, 
however, than the same amount of over-expansion, meaning 
that a nozzle that is too small for the required expansion is more 

* According to Dr. O. Recke, if the total divergence of a nozzle is more than 
6 degrees, eddies will begin to form in the jet. There is no doubt that a too rapid 
divergence produces a velocity loss. 

f Moyer's Steam Turbines. 



174 



FLOW OF FLUIDS 



efficient than one that is correspondingly too large.* Fig. 62 
shows a curve representing average values of nozzle loss | used 
to determine discharge velocities from nozzles under the condi- 
tions of under- or over-expansion. 

Non-expanding Nozzles. All the nozzles of Rateau steam 
turbines and usually also those of the low-pressure stages of Curtis 
turbines are made non-expanding; meaning, that they have the 
same area at the throat as at the mouth. For such conditions 
it has been suggested that instead of a series of separate nozzles 
in a row a single long nozzle might be used of which the sides 




.*-■*! 1(U9 





J6.19£H 

Fig. 63. — Non-expanding Nozzles. 



were arcs of circles corresponding to the inside and outside pitch 
diameters of the blades. Advantages would be secured both on 
account of cheapness of construction and because a large amount 
of friction against the sides of nozzles would be eliminated by 
omitting a number of nozzle walls. Such a construction has not 
proved desirable, because by this method no well-formed jets 
are secured and the loss from eddies is excessive. The general 
statement may be made that the throat of a well-designed nozzle 
should have a nearly symmetrical shape, as for example a circle, 
a square, etc., rather than such shapes as ellipses and long rect- 

* It is a very good method to design nozzles so that at the rated capacity the 
nozzles under-expand at least 10 per cent, and maybe 20 per cent. The loss for 
these conditions is insignificant, and the nozzles can be run for a large overload 
(with increased pressures) in nearly all types without immediately reducing the 
efficiency very much. 

f C. P. Steinmetz, Proc. Am. Soc. Mech. Engineers, May, 1908, page 628; J. 
A. Moyer, Steam Turbines. 



MATERIALS FOR NOZZLES 1 75 

angles. The shape of the mouth is not important. In Curtis 
turbines an approximately rectangular mouth is used because 
the nozzles are placed close together (usually in a nozzle plate 
like Fig. 60) in order to produce a continuous band of steam; 
and, of course, by using a section that is rectangular rather than 
circular or elliptical, a band of steam of more nearly uniform 
velocity and density is secured. 

Fig. 63 shows a number of designs of non-expanding nozzles 
used by Professor Rateau. The length of such nozzles beyond 
the throat is practically negligible. Curtis non-expanding noz- 
zles are usually made the same length as if expanding and the 
length is determined by the throat area. 

Materials for Nozzles. Nozzles for saturated or slightly super- 
heated steam are usually made of bronze. Gun metal, zinc 
alloys, and delta metal are also frequently used. All these 
metals have unusual resistance for erosion or corrosion from the 
use of wet steam. Because of this property as well as for the 
reason that they are easily worked with hand tools * they are 
very suitable materials for the manufacture of steam turbine 
nozzles. Superheated steam, however, rapidly erodes all these 
alloys and also greatly reduces the tensile strength. For nozzles 
to be used with highly superheated steam, cast iron is generally 
used, and except that it corrodes so readily is a very satisfactory 
material. Commercial copper (about 98 per cent) is said to 
have been used with a fair degree of success with high super- 
heats; but for such conditions its tensile strength is very low. 
Steel and cupro-nickel (8 Cu + 2 Ni) are also suitable materials, 
and the latter has the advantage of being practically non-cor- 
rodible. 

The most important part of the design of a nozzle is the deter- 
mination of the areas of the various sections — especially the 
smallest section, if the nozzle is of an expanding or diverging 
type. In order to calculate the areas of nozzles we must know 
how to determine the quantity of steam (flow) per unit of 
time passing through a unit area. It is very essential that the 
* Nozzles of irregular shapes are usually filed by hand to the exact size. 



176 FLOW OF FLUIDS 

nozzle is well rounded on the " entrance " side and that sharp 
edges along the path of the steam are avoided. Otherwise it 
is not important whether the shape of the section is circular, 
elliptical, or rectangular with rounded corners. 

Whether the nozzle section is throughout circular, square, or 
rectangular (if these last sections have rounded corners) the 
efficiency as measured by the velocity will be about 96 to 97 per 
cent, corresponding to an equivalent energy efficiency of 92 to 94 
per cent. 

PROBLEMS 

1. Air at a temperature of ioo° F. and pressure of 100 lbs. per sq. in. ab- 
solute flows through a nozzle against a back pressure of 20 lbs. per sq. in. 
absolute. Assuming the initial velocity to be zero, what will be the ve- 
locity of discharge? 

2. If the area at the mouth of the above nozzle is 0.0025 sq. ft. and the 
coefficient of discharge is unity, how many pounds of air will be discharged 
per minute? 

3. What will be the theoretical kinetic energy per minute of the above 
jet assuming no frictional losses? 

4. Steam at a pressure of 150 lbs. per sq. in. absolute and 3 per cent 
moisture flows through a nozzle against a back pressure of 17 lbs. per sq. in. 
absolute. Calculate the velocity at the throat of the nozzle. What will 
be the velocity at the end of the nozzle? 

5. Steam at a pressure of 200 lbs. per sq. in. absolute, temperature 
530 F., expands in a nozzle to a vacuum of 28.5 ins. (30-in. barometer). 
Calculate the absolute velocity of the steam leaving the nozzle. 

6. The area of a nozzle at its smallest section is 0.75 sq. in. It is sup- 
plied with dry saturated steam at 160 lbs. per sq. in. absolute pressure and 
the exhaust pressure is 2 lbs. per sq. in. absolute. How many pounds of 
steam per hour will the nozzle discharge? 

7. A steam nozzle is to be designed that will discharge 500 lbs. of steam 
per hour. The pressure of the steam is 175 lbs. per sq. in. absolute and has 
ioo° F. superheat. The exhaust pressure is 28 ins. vacuum (barometer 
29.82). What will be the area of the nozzle at its smallest section and at its 
end? 

8. A safety valve is to be designed for a 200 horse power boiler generating 
steam at 150 lbs. per sq. in. absolute pressure and 5 per cent moisture. 
Assuming that the safety valve should have a capacity such that it will 
release the boiler of all steam when generating double its rated capacity, 
what will be the smallest sectional area of the valve? 



PROBLEMS 177 

9. An injector supplies water to a boiler. The boiler pressure is 100 lbs. 
per sq. in. absolute and the steam generated is dry and saturated, the tem- 
perature of the entering feed water is 6o° F. and the temperature of the dis- 
charged mixture of condensed steam and feed water is 180 F. The suction 
head is 3 ft. What is the weight of feed water supplied to the boiler per 
pound of steam? What is the mechanical efficiency of the injector? 

10. Design a nozzle to deliver 400 lbs. of steam per hour, initial pressure 
175 lbs. per sq. in. absolute, final pressure atmospheric (barometer 28.62 
ins.), temperature of steam 6oo° F. 



CHAPTER X 

APPLICATIONS OF THERMODYNAMICS TO COMPRESSED AIR AND 
REFRIGERATING MACHINERY 

COMPRESSED AIR 

Air when compressed may be used as the working medium in 
an engine, in exactly the same way as steam. Furthermore, it 
is an agent for the transmission of power and can be distributed 
very easily from a central station for the purpose of driving 
engines, operating quarry drills and various other pneumatic 
tools. 

Air Compressors. The type of machine used for the compres- 
sion of air is that known as a piston-compressor and consists of 
a cylinder provided with valves and a piston. 

The work performed in the air-cylinder of a compressor can 
best be studied successfully from an indicator diagram. If the 
compression is performed very slowly in a conducting cylinder, 
so that the air within may lose heat by conduction to the atmos- 
phere as fast as heat is generated by compression, the process 
will in that case be isothermal, at the temperature of the atmos- 
phere. Also if the compressed air is distributed to be used in 
compressed air motors * or engines without a change of temper- 
ature, and that the process of expansion in the compressed air 
motors or engines is also indefinitely slow and consequently 
isothermal, then (if we neglect the losses caused by friction in 
pipes) there would be no waste of power in the whole process of 
transmission. The indicator diagram would then be the same 
per pound of air in the compressor as in the air motor or engine, 
although the course of the cycle would be the reverse — that is, 
it would retrace itself. 

* Compressed air motors are similar to steam engines, but use compressed air 
instead of steam. 

178 



WORK OF COMPRESSION 



179 



Adiabatic compression and expansion take place approximately 
if the expansion and compression are performed very quickly, 
or when the air is not cooled during compression, — in such 
a case the temperature of the air would rise. The theoretical 
indicator diagram of the compressor, Fig. 64, is FCBE and that 



3Y 





Volume 
Fig. 64. — Diagram of Compressor. 



Volume 
Fig. 65. — Diagram of Air Engine. 



of the air engine, Fig. 65, is EADF. CB and AD are both 
adiabatic lines. The change of volume of the compressed air 
from that of EB to EA occurs through its cooling in the dis- 
tributing pipes, from the temperature produced by adiabatic 
compression down to the temperature of the atmosphere. 

Suppose both diagrams of the compressor and of the air engine 
are superimposed as in Fig. 66, and then an imaginary isothermal 
line is drawn between the points A and C. 



P 




D 



Volume 
FlG. 66. — Superimposed Diagrams of Figs. 21 and 22. 

It is then evident that the use of adiabatic compression causes 
a waste of power which is measured by the area ABC, while the 
use of the adiabatic expansion in the air engine involves a further 
waste, shown by the area ACD. 

Work of Compression. Assuming no clearance in the com- 
pressor and isothermal compression, the pressure-volume diagram 
will be similar to Fig. 64. The work of the cycle will be 



l8o COMPRESSED AIR AND REFRIGERATING MACHINERY 



W = PcVc ~ PcVdoge^ 

V b 



PbV 



bV b, 



which becomes, since P C V, 
W = 



P b V b , 



Vc 

Vb 



\ -PcVAoge^. 

V c 



(236) 



(237) 



In practice the compression cannot be made strictly isother- 
mal, as the operation of the piston would be too slow. The dif- 
ference between isothermal and adiabatic compression (and ex- 
pansion) can be shown graphically as in Figs. 67 and 68. In 





Volume 
Fig. 67. — Compression Diagram. 



Volume 
Fig. 68. — Expansion Diagram. 



these illustrations the terminal points are correctly placed for a 
certain ratio for both compression and expansion. Note that in 
the compressing diagram (Fig. 67) the area between the two 
curves, ABC, represents the work lost in compressing due to heat- 
ing, and the area between the two curves, ACMNF (in Fig. 68), 
shows the work lost by cooling during the expansion. The 
isothermal curve AC will be the same for both cases. 

The temperature of the air is prevented as far as possible 
from rising during the compression by injecting water into the 
compressing cylinder, and in this way the compression curve will 
change. The curves which would have been PV = a constant, if 
isothermal, and PV 14 = a constant, if adiabatic, will be very much 
modified. In perfectly adiabatic conditions the exponent u n" 
= 1 .40 for air, but in practice the compressor cylinders are water- 
jacketed, and thereby part of the heat of compression is con- 
ducted away, so that " n" becomes less than 1.40. This value 
of " n" varies with conditions; generally the value is between 
1.2 and 1.3. 



EFFECT OF CLEARANCE UPON VOLUMETRIC EFFICIENCY 181 

When the compression curve follows the law, PV n equals a 
constant, work of compression (Fig. 64) is 

w = p c v c + PcVc ~ PbVb - P b V b 

n — 1 

(PcVc - P b V b ), (238) 



n — 1 



1 



since 



P C V» = P»V»*; V b = V c (Q" (239) 

Combining equations (238) and (239), 




W-^P.V.[i-[^) J- (240) 

The Effect of Clearance upon Volumetric Efficiency. It is 

impossible to construct a compressor without clearance, con- 
sequently the indicator diagram differs from the ideal case. At 
the end of the discharge stroke, the clearance volume is rilled with 
compressed air. When the piston moves on its outward stroke, 
the clearance air expands and the suction valves of the compres- 
sor will be held shut until the piston has moved a sufficient dis- 
tance to permit the entrapped air to expand to atmospheric 
pressure. When that point is reached any further movement of 
the piston opens the suction valves and external air is drawn into 
the cylinder. Thus the entire stroke of the compressor piston 
is not effective in pumping air. The ratio of the volume of air 
pumped to the volume swept by the piston, or piston displace- 
ment of the cylinder, is termed volumetric efficiency. 

Fig. 69 illustrates an ideal compressor diagram with clearance. 
The air entrapped in the clearance space equals 

v 3 = CV S , 

where V s = volume swept or piston displacement of the cylinder. 
C = percentage of clearance. 

The air in Fig. 69 expands to F 4 at which point the inlet valves 
open. The air drawn into the cylinder is represented by the 



182 COMPRESSED AIR AND REFRIGERATING MACHINERY 



difference in volume between Vi and F 4 . This air, as well as the 
clearance air, is compressed to point 2, while the compressed air 
is discharged from points 2 to 3. Knowing the per cent of clear- 



ex, 



v 3 =cy s 










L I 3 
ft 




a\ 




i\ 
1 \ 

1 ^ 
1 

1 

1 
1 
1 
1 


v% 


\4 


\w 


^^-— _ 1 


i< 1 — 

1 
u 


v* 


1 
1 


\f 


1 

t 
J 


f*— 






v s 


1 



Volume 
Fig. 69. — Ideal Air Compressor with Clearance. 

ance, the volume swept by the piston (V s ), and the initial and 
final pressure, the volumetric efficiency (Vet) may be deter- 
mined from the following equations: 

P% (CV s ) n = PiVi". 

1 

V* = (£)CV S , 

since V x = V s + CV S , 

V 1 - V, = V s + CV S 



-®y 



CV S 



i\ -1 



i + Cii - 



p s v 



V s . 



TWO STAGE COMPRESSION 



183 



Therefore volumetric efficiency is 



Vet. = 




V s 



= i+C 



1-1 



i-m 



(241) 



The volumetric efficiency also decreases with the altitude, as 
the weight of a cubic foot of air decreases as the altitude increases. 

Two Stage Compression. The problem of economy, obviously,' 
becomes one of abstracting the heat generated in the air during 
the process of compression. As previously mentioned, this is 
partially accomplished by water-jacketing the cylinders, and also 
by water injection. Nevertheless, owing to the short interval 
within which the compression takes place, and the comparatively 
small volume of air actually in contact with the cylinder walls, 
very little cooling really occurs. The practical impossibility of 
proper cooling to prevent waste of energy leads to the alternative 
of discharging air from one cylinder after partial compression has 
been effected, into a so-called inter-cooler, intended to absorb 
the heat generated during the first compression, and then 
compressing the air to the final pressure in another cylinder. 




Volume 
Fig. 70. — Indicator Diagram of Two-stage Air Compressor. 

This operation is termed " two-stage " compression and when 
repeated one or more times for high pressures the term " multi- 
stage " compression applies. 

Referring to Fig. 70 and assuming the compression in a two- 
stage compressor to be adiabatic for each cylinder, the compres- 



184 COMPRESSED AIR AND REFRIGERATING MACHINERY 



sion curve is represented by the broken line ABDE ; the compres- 
sion proceeds adiabatically in the first or low-pressure cylinder 
to B; the air is then taken to a cooler and cooled under practi- 
cally constant pressure until its initial temperature is almost 
reached, and its volume reduced from HB to HD; it is then 
introduced to the second or high-pressure cylinder and com- 
pressed adiabatically along the line DE to the final pressure 
condition that was desired. It is seen that the compression 
curve approaches the isothermal line FA.* The isothermal con- 
dition is obviously desired and, in consequence, air-machines 
are built to approach that condition as nearly as possible. 

Referring to Fig. 70, the work of each stage from equation (240) 
will be 



W (1st stage) = 



W (2d stage) 



n 
n — 



n 



r p °4 - ffi) " )• 



The total work of compression is 

W (total) = W (1st stage) + W (2d stage) 



n 



n 



PaVa 



1—^7- 



n—V 



+ PaV c 



B-n 



©' 



With perfect cooling P a V a = P d V a ; also P 6 = P d ; P c = Pe, then 

n— 1 / n— 1\ "" 



W (total) 



n 



n 



-PaVa 



<rH-(rj-<- 



The work of compression as expressed by equation (242) be- 
comes a minimum when 

n— 1 «— 1 



© " + (8) " 



(243) 



is a maximum. Since the initial and final pressures P a and P c 

are fixed, the pressure in the receiver, P b , can be found by differ- 

* The line FE represents further cooling. 



REFRIGERATING MACHINES OR HEAT PUMPS 



185 



entiating equation (243) and equating this differential to zero, or 



d 



dPi 



or 



n-l 



n-l-l 



m'Hpyi 



1 — n\ Pi 



n 



n 



n 



l-n 



ft 



l-n 



= 0, 



,Pc n 




/ "— 1 1 — 2» \ 
* LlLl (Pc n P b n ) 



.-. P & 2 = PoPc (244) 

For compressing air to high pressures three and four stage com- 
pressors are used. 

REFRIGERATING MACHINERY 

Refrigerating Machines or Heat Pumps. By a refrigerating 
machine or heat pump is meant a machine which will carry heat 
from a cold to a hotter body.* This, as the second law of 
thermodynamics asserts, cannot be done by a self-acting proc- 
ess, but it can be done by the expenditure of mechanical work. 
Any heat engine will serve as a refrigerating machine if it be forced 
to trace its indicator diagram backward, so that the area of the 




Volume 
Fig. 71. — Pressure- volume Diagram of Carnot Cycle. 

diagram represents work spent on, instead of done by, the 
working substance. Heat is then taken in from the cold body 
and heat is rejected to the hot body. 

* This statement is not at variance with our knowledge that heat does not 
flow of itself from a cold body to a hotter body. 



1 86 COMPRESSED AIR AND REFRIGERATING MACHINERY 

Take the Carnot cycle, using air as working substance (Fig. 
71), and let the cycle be performed in the order dcba, so that the 
area of the diagram is negative, and represents work spent upon 
the working substance. In the stage dc, which is isothermal 
expansion in contact with the cold body R (as in Fig. 7, page 41), 
the air takes in a quantity of heat from R equal to wRT 2 log e r 
[equation (46)], and in stage ba it gives out to the hot body H a 
quantity of heat equal to wRT\ log e r. There is no transfer of 
heat in stages cb and ad. Thus R, the cold body, is constantly 
being drawn upon for heat and can therefore be maintained at a 
temperature lower than its surroundings. In an actual refriger- 
ating machine operating with air, the cold body R consists of a 
coil of pipe through which brine circulates while " working " air 
is brought into contact with the outside of the pipe. The brine 
is kept, by the action of the machine, at a temperature below 
3 2 F. and is used in its turn to extract heat by conduction from 
the water which is to be frozen to make ice. The " cooler " H, 
which is the relatively hot body, is kept at as low a temperature 
as possible by means of circulating water, which absorbs the heat 
rejected to H by the " working " air. 

The size of an air refrigerating machine is very large as com- 
pared with its performance. The use of a regenerator, as in 
Stirling's engine (Fig. 8), may be resorted to in place of the 
two adiabatic stages in the Carnot cycle, with the advantage of 
making the machine much less bulky. Refrigerating machines 
using air as working substance, with a regenerator, were intro- 
duced by Dr. A. C. Kirk and have been widely used.* The 
working air is completely enclosed, which allows it to be 
in a compressed state throughout, so that even its lowest 
pressure is much above that of the atmosphere. This makes 
a greater mass of air pass through the cycle in each revolution 
of the machine, and hence increases the performance of a 
machine of given size. In all air refrigerating machines the 

* See Kirk, On the Mechanical Production of Cold, Proc. Inst, of C. E., vol. 
XXXVII, 1874. Also lectures on Heat and its Mechanical Applications, in the 
same proceedings for 1884. 



SYSTEMS OF MECHANICAL REFRIGERATION 187 

temperature range must be high to produce a given refrigerating 
effect. 

In another class of refrigerating machines the working sub- 
stance, instead of being air, consists of a liquid and its vapor, 
and the action proceeds by alternate evaporation under a low 
pressure and condensation under a relatively high pressure. A 
liquid must be chosen which evaporates at the lower extreme of 
temperature under a pressure which is not so low as to make 
the bulk of the engine excessive. Ammonia, ether, sulphurous 
acid, and other volatile liquids have been used. Ether machines 
are inconveniently bulky and cannot be used to produce intense 
cold, for the pressure of that vapor is only about 1.3 pounds per 
square inch at 4 F., and to make it evaporate at any tempera- 
ture nearly as low as this would require the cylinder to be exces- 
sively large in proportion to the performance. This would not 
only make the machine clumsy and costly, but would involve 
much waste of power in mechanical friction. The tendency of 
the air outside to leak into the machine is another practical ob- 
jection to the use of so low a pressure. With ammonia a dis- 
tinctly lower limit of temperature is practicable: the pressures 
are rather high and the apparatus is compact. Carbonic acid 
has been used as a refrigerant in small machines. The objection 
to carbonic acid is that the pressures are very high as compared 
with ammonia (see Appendix). The critical temperature of 
carbonic acid is less than 90 F. 

Unit of Refrigeration. The capacity of a refrigerating ma- 
chine is usually expressed in tons of refrigeration or ice melting 
effect per twenty-four hours. As the latent heat of fusion of 
ice is about 144 B.t.u., the heat units withdrawn per ton of re- 
frigerating effect per twenty-four hours is 

2000 X 144 = 288,000 B.t.u. . 

Systems of Mechanical Refrigeration. The standard systems 
of mechanical refrigeration are:* 

(A) The dense-air system, so-called because the air which is 

* Lucke's Engineering Thermodynamics, page 1148. 



1 88 COMPRESSED AIR AND REFRIGERATING MACHINERY 



the medium is never allowed to fall to atmospheric pressure, so 
as to reduce the size of the cylinders and pipes through which a 
given weight is circulating. 

(B) The compression system, using ammonia, carbon dioxide 
or sulphur dioxide, and so-called to distinguish it from the third 



A 



I 



Circulating J 
Water I 





g^ 




A 












S 










Cool Com- 
pressed Air 


f 






Hot Com- 
pressed Air 

A 








, 


^ 


^ 










Expt 
Cyl 


<? . 




F 

Compressor 

Cylinder 




E 
Engine 
Cylinder 






nder 












! 




j 


\ 










> 

Cold Low 
Pressure Air 








Warmed Low 
Pressure Air 










/ 








" \ 


V 




1 < B 






- 




• 


1 ' 






Brine Tank 






1 ~ 








" j* 





[Circulating Brine] 
Warm i Pipes \lcold 

j jBrine 



Brine 



Fig. 72. — Dense Air System of Refrigeration. 

system, because a compressor is used to raise the pressure of the 
vapor and deliver it to the condenser after removing it from the 
evaporator. 

(C) The absorption system, using ammonia, and so-called be- 



THE AIR SYSTEM OF REFRIGERATION 189 

cause a weak water solution removes vapor from the evaporator 
by absorption, the richer aqua ammonia so formed being pumped 
into a high-pressure chamber called a generator in communica- 
tion with the condenser, where the ammonia is discharged from 
the liquid solution to the condenser by heating the generator, 
to which the solution is delivered by the pump. 

No matter what system is used, circulating water is employed 
to receive the heat, the temperature of which limits the highest 
temperature allowable in the system and indirectly the highest 
pressure. 

The Air System of Refrigeration. The dense or closed air 
system is illustrated in Fig. 72, in which air, previously freed of 
moisture, is continuously circulated. The engine cylinder E 
furnishes power * to drive the compressor cylinder F. This 
cylinder delivers hot-compressed air into a cooler A where it 
is cooled, and then passed on to the expansion cylinder G (tan- 
dem-connected to both, the compressor F and to the engine 
cylinder E), which in turn sends cold low-pressure air first through 
the refrigerating coils in the brine tank 
B and then back to the compressor cylin- 
der F; thus the air cycle is completed. 
The courses of the circulating water and 
also of the brine are shown by the dotted 
lines. 

The dense-air cycle in a pressure- volume Fig. 73. — Pressure- vol- 
diagram is represented in Fig. 73, in which um e Diagram of Dense Air 
BC is the delivered volume of hot-corn- Cycle of Ref ^re- 
pressed air; CM is the volume of cooled air admitted in the ex- 
pansion cylinder; MB the reduction in volume due to the water 
cooler; MN the expansion; NA the refrigeration or heating of 
the air by the brme, and AB the compression. This operation is 
but a reproduction of that previously described. 

The work W c expended in the compression cylinder F is DABC 

* Since the work done by the expansion of the cool-compressed air is less than 
that necessary for the compressing of the air taken from the brine coils through 
the same pressure conditions, a means must be employed to make up for the dif- 
ference, and for this purpose the engine cylinder is used. 




190 COMPRESSED AIR AND REFRIGERATING MACHINERY 

(Fig. 73) ; that done by the expansion cylinder G is W e = DCMN. 
The shaded area MB AN represents the work which must be sup- 
plied by the engine E. 

If w pounds of air are passing through the refrigerating machine 
per minute, the heat withdrawn from the cold room or absorbed 
by the brine along NA (Fig. 73) is 

Qna = wC P (t a — h). (245) 

The work expended in compressing w pounds of air, assuming 
poly tropic compression, is, by equation (240), 

If the compression is adiabatic, 



W c =wP a Va-^ L - 

7 - I 



since -=?- = ( — ) v also AR = C v — C v = C v I- J 

If the expansion is complete in the expansion cylinder, the work 
done in expansion is 

We=^(t m '-h). (247) 

The net work of the engine E required to produce refrigeration 
becomes 

W = Wc - We = ^ (h ~ t a - t m + *„). (248) 

Since, from equation (245), 

w = -77^ (ta — t n ), equation (248) becomes 



l/*& ~~ ta — tm ~j~ tn\ 
\ t a -t n / 



7—7 \/NA[tb — t a — tm ~T t n \ 

W = j-( : — ; ) * ( 2 49) 



THE VAPOR COMPRESSION SYSTEM OF REFRIGERATION 1 91 

The Vapor Compression System of Refrigeration. The com- 
pression system for ammonia or similar condensable vapors is 
shown in Fig. 74. The figure illustrates the essential mem- 
bers of a complete compression refrigerating system. B repre- 



High Pres-> ' 
Bure Liquid 



R 
Ammonia 
Receiver 



Circulating 

Y Water 

I 



w 

Condenser 



<■ 



->- 



High Pres- 
Asure Vapor 



Compressor 
Cylinder 



E 

Engine 
Cylinder 




Throttling or 
Expansion Valve 

Low Pressure Liquid 



Low Pres- 
sure Vapor 



t= — Circulating Brine- 

J Pipes I 

Warm f \ Cold 

Brine I j Brine 

Fig. 74. — Compression System of Refrigeration. 



sents the direct-expansion coil in which the working medium is 
evaporated; F, the compressor or pump, for increasing the pres- 
sure of the gasified ammonia; E, the engine cylinder, — the source 
of power; W, the condenser, for cooling and liquefying the gasi- 
fied ammonia; and V, a throttling valve, by which the flow of 



192 COMPRESSED AIR AND REFRIGERATING MACHINERY 

liquefied ammonia under the condenser pressure is controlled as 
it flows from the receiver R to the expansion coils; B, the brine 
tank, in which a materially lower pressure is maintained by the 
pump or compressor in order that the working medium may boil 
at a sufficiently low temperature to take heat from and con- 
sequently refrigerate the brine which is already cooled. 

The operation of the compressor F (Fig. 74) is theoretically a 
reversed Rankine cycle (Figs. 31 and 32). If one pound of vapor 
is passed through the system, the work (Wc) of the compressor is 

W c = (#2 - #1) (B.t.u.), 
where 

Hi = the total heat of the vapor at entrance to the com- 
pressor, 

H 2 = the total heat of the vapor discharged from the com- 
pressor. 

The vapor entering the compressor may be treated as though 
it were dry and saturated. The entropy-temperature diagram 
(Fig. 75) illustrates this case. The vapor discharged by the 
compressor will then be superheated, having a temperature 
t s at a pressure P 2 . The condition of the discharged vapor is 
determined by equating the entropies at the inlet and discharge 
pressures. The total heat (H 2 ) of the discharged vapor will be 

H 2 = [fa + L 2 + C P (t s - k)]. (250) 

The horse power of the compressor is, if w pounds of vapor is 
circulated per minute: 

_ ^-#0 778, ( j 

33,000 

The heat absorbed per minute by w pounds of vapor passing 
through the refrigerating or brine coils will be 

Q r = w (fa + Li - fa) 

= w (Hi - fa). (252) 

fa = the liquid heat of the vapor after leaving 
the condenser. 



THE VAPOR ABSORPTION SYSTEM OF REFRIGERATION 193 



The heat absorbed per minute by the condenser (Q c ) is 

w (H 2 - ho) (B.t.u.). (253) 

The heat absorbed by the condenser is theoretically equal to 
the sum of the heat absorbed in the refrigerator and the heat 
equivalent of the work of compression. 

Ammonia compressors are operated either on the dry or on 
the wet system. In the dry system the ammonia entering the 
compressor is a dry vapor as illustrated in Fig. 75. In the wet 
system the ammonia enters the compressor in the wet state, the 
heat developed during the com- 
pression being used in evapo- 
rating the liquid ammonia into 
a vapor. 

The Vapor Absorption Sys- 
tem of Refrigeration. The ab- 
sorption system depends upon 
the fact that anhydrous am- 
monia possesses the property 
of forming aqua ammonia. 
The amount of ammonia water 
will absorb depends upon the 
temperature of the water; the 
colder the water, the greater are 
its absorptive powers. 

The absorption system differs 
from the vapor compression 
system in that the absorption system replaces the compressor 
by an absorber, where the anhydrous ammonia is changed into 
aqua ammonia, a pump which transfers the ammonia from the 
absorber to the generator, and a generator, where the aqua 
ammonia is heated. Both systems have a condenser where the 
ammonia is cooled and liquefied and an expansion valve or 
throttling valve by means of which the flow of liquid ammonia 
to the expansion coils is controlled. In the absorption system 
the anhydrous ammonia vapor flows from the expansion coils to 



T 2 
T/' 


/ p 2 y 


/ , \ 




Si 

S 

a 
£ 

H 







Fig. 75. 



Entropy -0 
— Entropy -Temperature Dia- 
gram for Ammonia. 



194 COMPRESSED AIR AND REFRIGERATING MACHINERY 

the absorber, in which the anhydrous vapor comes in contact 
with weak aqua ammonia. The weak ammonia absorbs the 
ammonia vapor. From the absorber the ammonia is pumped 
into the generator, where it is heated by steam coils. The vapor 
driven off in the generator passes to the condenser and from there 
through the expansion or throttle valve to the expansion coils 
which are located in the brine tank or in the refrigerating room. 
The absorption system is usually provided with a rectifier to 
thoroughly dry the gas before it enters the condenser, an ex- 
changer which heats the strong ammonia by cooling the weak 
ammonia, and with other auxiliary equipment to reduce the heat 
losses in the system. 

COEFFICIENT OF PERFORMANCE OF REFRIGERATING MACHINES 

The Coefficient of Per- 1 ___ Heat extracted from the cold body 
formance j" Work expended 

This ratio may be taken as a coefficient of performance in esti- 
mating the merits of a refrigerating machine. When the limits 
of temperature 2\ and T 2 are assigned it is very easy to show by 
a slight variation of the argument used in Chapter IV that no 
refrigerating machine can have a higher coefficient of performance 
than one. which is reversible according to the Carnot method. 
For let a refrigerating machine S be driven by another R which 
is reversible and is used as a heat-engine in driving S. Then if S 
had a higher coefficient of performance than R it would take from 
the cold body more heat than R (working reversed) rejects to 
the cold body, and hence the double machine, although purely 
self-acting, would go on extracting heat from the cold body in 
violation of the Second Law of Thermodynamics. Reversi- 
bility, then, is the test of perfection in a refrigerating machine 
just as it is in a heat-engine. 

When a reversible refrigerating machine takes in all its heat, 
namely Q c at T 2j and rejects all, namely Q a at T\, and if we repre- 
sent the heat equivalent of the work done by W = Q a — Qc, then 
the coefficient of performance is, as already denned, 



PROBLEMS 195 

Qc _ Qc T 2 



W Qa-Qc ZV-T, 



(254) 



Hence, the smaller the range of temperature, the better is the 
performance. To cool a large mass of any substance through a 
few degrees will require much less expenditure of energy than to 
cool one-fifth of the mass through five times as many degrees, 
although the amount of heat extracted is the same in both 
cases. If we wish to cool a large quantity of a substance it is 
better to do this by the direct action of a refrigerating machine 
working through the desired range of temperature, than to cool 
a portion through a wider range and then let this mix with the 
rest. This is only another instance of a wide, general principle, 
that any mixture of substances at different temperatures is ther- 
modynamically wasteful because the interchange of heat between 
them is irreversible. An ice-making machine, for example, 
should have its lower limit of temperature only so much lower 
than 3 2 F. as will allow heat to be conducted to the working 
fluid with sufficient rapidity from the water that is to be frozen. 

PROBLEMS 

1. If 200 cu. ft. of free air per minute (sea-level) is compressed isother- 
mally and then delivered into a receiver, the internal pressure of which is 
102.9 lbs. per sq. in. absolute, find the theoretical horse power required. 

2. What will be the net work done in foot-pounds per stroke by an air 
compressor displacing 3 cu. ft. per stroke and compressing air from atmos- 
pheric pressure to a gage pressure of 75 lbs.? (Isothermal compression.) 

3. What horse power will be needed to compress adiabatically 1500 
cu. ft. of free air per minute to a gage pressure of 58.8 lbs., when n equals 

4. A compressed-air motor without clearance takes air at a condition 
of 200 lbs. per sq. in. (gage) and operates under a cut-off at one-fourth 
stroke. What is the work in foot-pounds that can be obtained per cubic 
foot of compressed air, assuming free air pressure of 14.5 lbs. and n equal 
to 1 .41? 

5. Find the theoretical horse power developed by 3 cu. ft. of air per 
minute having a pressure of 200 lbs. per sq. in. absolute, if it is admitted 
and expanded in an air engine with one-fourth cut-off. The value of n is 
1.2. (Neglect clearance.) 



196 COMPRESSED AIR AND REFRIGERATING MACHINERY 

6. Compute the net saving in energy that is effected by compressing 
isothermally instead of adiabatically 50 cu. ft. of free air to a pressure of 
200 lbs. per sq. in. gage. Barometer = 14 lbs. per sq. in., and a tempera- 
ture of 70 F. What is the increase in intrinsic energy during each kind 
of compression? How much heat is lost to the jacket-water during each 
kind of compression? 

7. Let a volume of 12 cu. ft. of free air be adiabatically compressed in a 
one stage air compressor from atmospheric pressure (15 lbs.) to 85 lbs. gage; 
the initial temperature of the air being 70 F. 

(a) What is the volume and temperature of the air after the com- 

pression? 

(b) Suppose this heated and compressed air is cooled to an initial 

temperature of 6o° F. at constant volume, what is its pressure 
for that condition? 

(c) Now if the air occupies such a volume as found in (a) and at 

an absolute pressure as determined in (b), at a temperature 
of 6o° F., and is then allowed to expand adiabatically down 
to atmospheric pressure (15 lbs.), what is the temperature 
and volume of the expanded air in Fahrenheit degrees? 

8. Compare the work required to compress one pound of air from atmos- 
pheric pressure and a temperature of 6o° F. to a pressure of 200 lbs. per sq. 
in. absolute in (1) a one-stage and (2) in a two-stage compressor. Assume 
n = 1.25. 

9. Using equation 249 as a basis, deduce a formula for the horse power 
required to abstract a given number of heat units per minute by an air 
refrigerating machine. 

10. Calculate the approximate dimensions of a ten-ton air refrigerating 
machine and the power required to drive it. Pressure in the cold chamber 
is atmospheric and the temperature 32 F. Pressure of air delivered by 
compressor is 90 lbs. per sq. in. gage. The temperature of the air coming 
from condenser is 85 F. and the machine operates at 60 r.p.m. Allow 
12 lbs. drop in pressure between the compressor and the expanding cylinder. 

11. Calculate the approximate dimensions of a ten-ton ammonia com- 
pression refrigerating machine and the power required to drive it. The 
temperature in the expansion coils is 15 F. and the temperature in the con- 
denser is 8 5 F. The machine is double acting and operates at a speed of 
80 r.p.m. 

12. What is the ice-making capacity per twenty-four hours of the machine 
in problem 11? The temperature of the water to be frozen is 8o° F. 



APPENDIX 



Table I 

SPECIFIC HEAT OF GASES AND VAPORS 
(Taken from Smithsonian Physical Tables) 



Substance 



Air. 



Alcohol, 
(ethyl) 

Alcohol, 
(methyl) 



Ammonia . 



Benzene. 



Carbon 
Dioxide 



Carbon 
Monoxide 



Ether. 



Hydrogen. . 

Nitrogen 
Sulphur 
Dioxide 

Water 



Range of 
Temp.° C. 



-30-10 
0-100 
0-200 
20-100 
Mean 

108-220 

101-223 

23-100 
27-200 
24-216 

Mean 

34-iiS 

35-i8o 

116-218 

-28-77 

15-100 
11-214 

Mean 



23-99 

26-198 

69-224 
27-189 
25-1 1 1 
Mean 
-28-9 
12-198 
21 — 100 
Mean 
0-200 

16-202 
128-217 
100-125 

Mean 



Cv 



0.23771 

0.23741 

0.23751 

0.2389 

0.23788 

0-4534 

0.4580 

0.5202 
o.5356 
0.5125 

0.5228 

0.2990 
0.3325 
0-3754 

0.1843 

0.2025 
0.2169 

0.2012 



0.2425 
0.2426 



•4797 
.4618 
.4280 
■4565 
•3996 
.4090 
.4100 
.4062 
.2438 

■1544 
.4805 

3787 
4296 



Authority 



Regnault 
1 1 

1 1 
Wiedemann 

Regnault 

Regnault 
Wiedemann 

Regnault 

Wiedemann 
< < 

Regnault 

Regnault 



Wiedemann 



Regnault 
Wiedemann 



Regnault 

Wiedemann 

Regnault 

Regnault 

Regnault 

Gray, 
Macfarlane, 



Cv 
C V 



1 .4066 
i.i3 6 



131 



1.300 



1-403 



1 .029 



1 .410 
1 .410 

1 .26 



1.300 



Authority 



Various 
Jaeger, j 
Neyreneuf / 



Cazin, 
Wiillner 



Wiillner 
Rontgen, 



Cazin, 

Wiillner 



Muller 



Cazin 
Cazin 
Cazin 
Muller 



Various 



Calcu- 
lated 
Co 



O.1691 
O.3991 



O.3991 



O.1548 



O.1729 



O.4436 



2.419 
0.1729 

O.1225 



0.3305 



197 



198 



APPENDIX 



Table II 

DENSITY OF GASES 
(Taken from Smithsonian Physical Tables) 



Gas 


Specific 
gravity 


Grams per 
cubic centimeter 


Pounds per 
cubic foot 


Air , 

Ammonia 


I .OOO 

0.597 
I.529 
O.967 
O.320-O.740 
O.0696 
1 .191 

0.559 
O.972 
1 .105 
2.247 


O.OOI293 
O.OOO770 
O.OOI974 
O.OO1234 

. 000414-0 . 00095 7 
0.000090 
0.001476 
0.000727 
0.001257 
0.001430 
0.002785 


O.08071 
O.04807 
O.12323 
O.07704 
O.02583-O.05973 
0.00562 
O.09214 
0.04538 
O.07847 
O.08927 
O.17386 


Carbon dioxide. ...... 

Carbon monoxide 

Coal gas 


Hydrogen 


Hydrogen sulphide. . . 
Marsh gas 


Nitrogen 


Oxygen 


Sulphur dioxide 



APPENDIX 



199 



Density 
Weight per 
Cubic Foot, 

Pounds. 


3 


OO H OMOO PO W On 

rj- voO lOV)iO^- W)H O O von rrO n 10 w On O 
POO On cm 10OO m ^tf- N O N^tOvO hi O m\0 O "^O 
OOOhhihicmcmcmpo 1000 m POO 00 m POO 00 O 

OOOOOOOOOOOOHIMMHICMCMCMCMPO 

000000000000000000000 


000000000000000000000 


a! 

p. 


•» 


PO Ovo NO 00 O O 
OOOOOOOOOnOvovoio pOOO 10 cm co co m po 


lO^H tj- CM HI N CM N PO POOO O PO HI fONNOO lO CM 
tO pi 't Q> ^ ^O HO f)N h OnNNiO'^-'^-pOpOpO 
Omo O no ioi- ^-t^toH w 

CM HI H 


Si 


+ 


OO NIOO CM Tj- fO <M PO^OOO -TJ- CM lO HI POOO rj- r- t^~ 
N « NtO POOO 10 ^- -^- lOOO ^- M PO00 O IOICNOM 
t-^ M MOtOH O00 NHOOO ^j- CM M 000 N N 

mhiOOOOOOnOnOn Onoo 0000000000 j>-r^.r^r^ 


CMCMCMCMCMCMCMHIHIHIHIHIHIHIH1HHHIHIHIH1 


Entropy 
of the 
Vapor. 


k]I^ 


O r}- *0 O PO 10O NOO N h OO •^■■^•w O CM N 10O 
VO O POPOH lorCtNN CO Tj- M 00 M0000 O rr OO 

N.H n ^- h aNin^- ^too rj-Ooo 10 po cm 000 n 

w O O On On 000 00 00 00 r^O OO 10 10 10 10 10 "t ^|- 


CMCMCMHIHIHIHIHIHIHIHIHIH1HIHHIHIHIHHIH 


Entropy 

of the 

Liquid. 


<£> 


CM POO OO O^N !OlON O00 00 00 M O POO CM CM N 
O CM TJ- O CM CM HI OnO CM Tj- O Oi<tNNM^f)00 

O rJ-O OOOnOhihicmponOhipovo *0O NOO O On 

OOOOOHHIHIHII-IHlCMCMCMCMCMCMCMCMCMCM 


OOOOOOOOOOOOOOOOOOOOO 


Heat 

Equivalent 

of 

External 

Work. 




•^ 10O to co On rr O -+N POOO O O 00 tt O O w tnoo 


^O NOO OnOnOmwhtj-vo NOO 00 On O O H H H 
loioirjioio too OOOOOOOOO t^r^t^t^f>» 


Heat 

Equivalent 

of" 

Internal 

Work. 


a 


POM N ^VO On NOO N ©N'tO "3- O -^ CM -"tf- O00 00 


N lO N CM NfOO N to CM O O 00 CM n cm 00 tt O n -^- 

M Q On OnOO O000 NNNio^J-rOrOM cm m m m O O 
O On On On On On On On On On On O On On On On On On On On 

M HI 




+ 

II 

&5 


NOO OnO O M POCM O^OO IOIONIOO M H ON to 


•sj- CM N M <t N 0\H CM ^-ioh<3 O POO On H PO ^O 
NOO 00 OOnOnOnO O O w CM CM fOtOCOrO^ - ^-'<t'^ - 
OOOOOOOmmhhmmhhmmmhmh 
mmwmmhhhmmhhhmhmhhmmh 


Heat of 
Vaporiza- 
tion. 


►J 


NO POO O 00 w NO O O co I s * POOO OO CM O O cm O 


M M IOO N CO HI 00 O Tf HI CM IOO IOMOO tr> CM OnO 
f-~0 lOlOTj-T}--^POPOPOCM m O O On OnOO OOOO NN 
OOOOOOOOOOOOOOOnOnOnOnOnOOn 

HHMHHHMMHHMhHMM 


Heat of 

the 
Liquid. 


•« 


lOtOMOH TTOO tO H 

O CM to On N CO hi Tf POOO O 'ct - On h On NOO N h N a 


C)h N O N POOO HO On^OO O N -<3- O O h to On 
CM PO ^r ^" VO IOO O O "On O CM PO CO *3" 10 IOO 

MMMMMHMMHM 


Tempera- 
ture, 
Degrees F. 


- 


CO to On h 00 cmoOO co co to cm mOOO too N cm too 
O m tt OnO CO h <sf POOO hioOmOOOOOnmno 


10 PO •"*■ CM On >0 O "3-00 MO m pocm OO CMOO ro N H 
PO IOO N t>.00 On On On O CM tJ- 10O !>• r^OO 00 On On O 


Absolute 

Pressure, 

Pounds per 

Square 

Inch. 


■fc. 


h cm po ^f IOO N00 On 


OOOOOOOOOmcmpOtJ- 10O N00 0> O h cm 

H — ■!■ ■!■ H M M 



a 



200 



APPENDIX 



< 
ft 

co. 

P 
ft 
H 
< 
ft 

ft 
O 

CO 

ft 

i— i 

ft 
ft 
ft 
O 
ft 
ft 



3 

< 



Density 
Weight per 
Cubic Foot, 

Pounds. 


§ 


O 0\«\0 o 

POO toOW rrOO m ^ ifl io to »t (O N m O^t^'sJ-HOO 

PO to t^OO O^h N -t toO f^OO a O m N N POrj-iovo 

PO PO PO fO ^O t^OO OO H N W) toO t-»00 Os O m M 

OOOOOOOOOmi-iihmmmmmmw<n<n 

ooooooooooooooooooooo 




3 5 

ft O 


■a 


Os 
PO <N Q\ t-»00 O Tj- Os OS Os H 00 MOO w t>»0 Os to N 
O O **» <N O CO t^OO i- W5 10NHO MOO 't HOO^O ^- 




OOOOO O O pom O OsOO *•*• t^O O to to to tJ- rj- tJ> 

PONNNNMHHM 




el 


+ 


"t <t ID On O^O H00 M lOH to <N 00 t>- N O w to M O 
O O O ■* <N <0 OsO O O 00 O COM3 O to O toOO <N 
O O VO to PO HH OsOO NO lO to Tf- PO PO <N M M M O O 

r>» r-» N N r-» t>.vo OOOOOOOOOOOOOO 




MHMHMMHMHHtHHHMMMHHWI-tM 




Entropy 
of the 
Vapor. 


•^Ih 


Os po NO to ^t h O m •^-oo ON O TfO 00 to M M N N 
PO N >t h vO O hO rh rj-O O O PO Os NO O O O N 
so to tt tt OsO poOCOso -^pOih OOO NO to Tf PO N 

^-^r^-TrPOCOCOCONNCSNNNMMHHHMH 




Entropy 

of the 

Liquid. 


<*> 


lO HI OO f)>0« OOO O M POO N TfM TJ- lO O ^" ^- PO 

N 00 h PO to PO00 so <N N m Os n tj- m NcoO't Oi^ 

O O M H PO tOSO OO^OHHNtTt^lO tOSO SO N 




OOOOOOOOOOOOOOOOOOOOO 




Heat 

Equivalent 

of 

External 

Work. 


a 


NO Os ON PO <0 <N C\^0 M SO OS NOO H lo N O N ION 




« N <N csi rj- tosO so NOO OOOOOsOsOOOmmmm 

t>»tv.r^.t^t^t^r^t>»t>-t^t^.i>.r^ noo oo oo oo oo oo oo 




Heat 

Equivalent 

of 

Internal 

Work. 


a 


O <0O 000000 O w Os CO O NO rt-PO tOOO PO O NO 




<N Os t^.so tosO ON toO toO to m N co OsO PO OsO 
O Os OS OS 00 i^-soso iOto^-^-POPO<N N M M M O O 

OsOO oooooooooooooooooooooooooooooooooooooo 




's'i 


+ 
II 


O *3* Tf N <N ^- onoO ^J-so so rr O tooO 00 PO ^ tt ^ CO 




00 Os O O O O tOO Os m PO to t^OO Os hh csi po Tf toO 
rt- rj- to to toO OOO t^t^t^t>»r^ l>-00 OO 00 00 00 00 

WMHHHWHMMMHHMHMHIIHI-IHIMH 
HHHMHWMHH.HMMHHMMMHMMH 




Heat of 
Vaporiza- 
tion. 


hj 


M ON'Jtt^O O m OsPONtoO O^O N N(Oh OsOsO 




Tf H O Os O <N IO00 POOO POOsrl-M NtOO NtOOM 
t^ t^ t^-O Otorj-POPO<NCSHMMOOOOsOs OsOO 
OsOsOsOsOsOsOsOsOsOsOsOsOsOsOsOs O^OO 00 00 00 




Heat of 

the 
Liquid. 


i« 


00 to O O h ^-00 Os h rj- h PO h t^O ^ O PO to to PO 




PO t^ O m O 00 00 i>-0 POOO <N ion t^NO O Tf-OO 
t^. l>-00 OOOsOMNPO^to too O t*» t>-O0 OO Os Os Os 
HMHHMNNNNNNNNCNNNCSNNNCS 




Tempera- 
ture, 
Degrees F. 


- 


MOO 

OOtoO O O m POPOPOtoO h t^ O OsO ^0 pO O 00 




lO Os N POOO O O ON't H N N00 N t>-NO O ^ t"* 

O O h h n ^io toO t^oo OOOsOsOOmmnnn 

NNNNNNN<NCSC>lNN<NCSPOPO<OCOfOfOCO 


' 


Absolute 

Pressure, 

Pounds per 

Square 

Inch 


«. 


pO^rftoO toO toO toO toO toO toO too too 

MMMMCSNPOPO'^'^-tO tOO O t^ t^OO 00 Os Os O 

H 


L 



APPENDIX 20I 



to cn t^crjw r^. cn n fiO to o cooo cooo cn t^» cn 

CO t^ r^OO OOO O m <N CN CN to rO i" "t 10 lOO t*» r"» t^OO 00 On On O M rO 10 
\0 rf too r^OO O m <N co •>^- i^O t^OO O O m cn co ■>*• 100 t^OO O cn ■<*■ O tj- 
<n CN cn cn (N cn cocOPOcococococo<Oco-s}-T}-'!t-Tt'3--<3--<3-'3-Ti-Tj-toto toO 


o o o o o o 


OOOOOOOOOOOOOOOOOOOOOOOO 


o *■*» o o <o cn 
CO ^l-oo cn co to 
CN O oo j>- to ■<*■ 


m On cn cn O rf- co to cn CO00 O O O t^. r^OO H \0 TJ- rj- O O M 
com m m CN toiONO COO O Tt" On COOO co On tJ- O CN tOOO to 
co cn m O QnCO r-^O Oio^-^-cocncnmihOOO OnOO O to 


Tf ^ tO to to to 


COCOCOCOCNCN*CNCNCNCNCNCNCNCNCNCNCNCNCNCNMMMM 


O cn r^ co On r-* 
00 ^O Nt^O 
O O OnOO 00 00 
to to to to to to 


r^f^O\<N -^-CNtoO I^-coOOOOOOOOO Onm m OnO On On 
l^ -tJ- m OnO co m OnO ^-<N ONNiorOH On r-O "t O r~~ On CN 
t~~ r-^ t^-O OOO totototorJ--<^-Ti--rl--^-cococo<Oco<N M M 
totototototototototototototototototototototototo 






m 00 o ^- o o 
On O co tooo O 

w m O OnOO 00 

M M M O O O 


CN toCN O OnmnO hGO to'l-rtO OncOOC to m 0~0CO Q Onm 
•^- r^. m tooo cor^CNNO hO m no m t^CNOO -^-ONtor^O m to 
NOO io^--^-coco<N <N M M O O ON O00 00 t^-t^-OO "3" CN 
OOOOOOOOOOOOOOOnOnOnOnOnOnOnOnOnOn 


H 1-1 M M M M 


mmmmmmmmmmmmmmOOOOOOOOOO 


On 'sj- r~» On O00 ion no tooo o\ on OnOO O t O N cooo tOOO cn O coO O 00 
CO tON h lO 6fON O Tj-t^-O COVO On <N to00 hi COO O0 M COO 00 CO f^OO *>■ 
f>-00 0O O Os ON O O M M M CM <N (N <N tOf)tO^ - ^' , ti" 1 1/ 5 1 to O O t*» 00 
^rf^rt^rtlOlOlOtOtotOtotOtotOtototol-OtototOtotototovoiOlO 


o o o o o o 


OOOOOOOOOOOOOOOOOOOOOOOO 


On h co too 00 


On <N CO toO -t^OO On O H CN CO "^- to toO t>» t^OO O HI COO 


m CN cm CN cn cn cn cocOfOfOCOCTO(Oit , ti - i'i' , t't , t , J -, t^' l/ ' l O , O l O 
0OOOOOOO0O0OO00OO0COO0OOO00OOO00O00OO0O00O0OO0O0O0O0OO000O0O 


»0 N aMO O 


toO N^-HOOO iO'^-'!i- , ^-^-^J- too 00 O CO tOOO "cj- CN 00 


COO N lO N O MO N 000 to CO M O^N") fO H O^t^lO-^-CN O 00 tO<N Tj- O 

O O On O On OnOO 00 00 00 **» f>- f*~ r^NO OOOO totototototo-rtTj-TfcoCN 


cn 0000 cOOO cn 00 'd-O toO rj- On rt-oO f^NH tooo cm O On cn On tooo h 


t^00 00 On O m 
00 00 OO 00 On On 


m cn cn co^Tl-toto too O r*» t^-00 0000 O^ ON On O O m cn^ 

OnO^OnOnO^OnOnOnOnOnO^OnOnOnCnOnO^O^O^O o o o o 


CN tOOO N Nt<N 


OnO "<d- <n O 00 00 t^ r>-00 OO O O cn tJ-O On cn •rj-oo io ^OO co 


to cn ONt^^J"<N ONiorOHOOO tJ- cn OOOO to co m On t^-O ^- cn OnO 00 m 
OOOO NNls r^O OOOO tototototo^-Tj-Tj-'^- , ^-co<Oco<OcOCN) cn i-i m 

ooooooooooooooooooooooooooooooooooooocoooooooocoooooocoooooo 


O to,0 to too 


f^-O ^" CN ONO CN t^.CNNO O i-NONH CN •^■rJ-tOlO'chCN CN t^ 


CN to Cn <n tooo 

O O O M W M 

CO CO CO CO CO co 


H ^-NO « tOOO O CO tOOO O <N TfN O^H fO>ONH lOTfCN 
CN CN CN cOcO<OcO'^'^-^-'^-totototo toO O O O t*~ t^OO On 
cocococo<Ococococococococo<OCOcocococococotOcoco 


tJ-00 m CO ■* "*■ 


CO H 00 to O O O tOOO M •st-O 00 On O O On OnOO \tH V)ifl 


M tJ-OO h ^NO "5 tooo M COO OO O co to r^ On M ^O OO ONH !ON h ON t^ 

cococo-^-'^-'^-tototo too o o o i^ r^ t^ r-~ r^-co oooooooo a ONO>0 O m 
cococococococococococococococococo<OCOcococ r 5COcocococO , 5)"^t - ' , 4' 


to O to O to 

O M M CN CN CO 


toO <fl ioO too toO toO toO toO toO >oO O O toO 

CO Tf ^ to toO O f^ t~^O0 OOOiO>OOhhN(NI«5tj->ONO 







202 



APPENDIX 



Table IV. — PROPERTIES OF SUPERHEATED STEAM 

Reproduced by permission from Marks and Davis' "Steam Tables and Diagrams." 
(Copyright, 1909, by Longmans, Green & Co.) 



Pressure, 


Satu- 


Degrees of Superheat. 


Pressure, 


Pounds 


rated 




Pounds 


Absolute. 


Steam. 


50 


100 


150 


200 


250 


300 


Absolute. 




t 


162.3 


212.3 


262 .3 


312.3 


362.3 


412.3 


462.3 


/ 




5 


V 


73-3 


79-7 


85-7 


91.8 


97-8 


103.8 


109.8 


V 


5 




h 


"30- 5 


H53-5 


1176.4 


H99-5 


1222 .5 


1245.6 


1268.7 


h 






t 


193.2 


243.2 


293.2 


343-2 


393-2 


443-2 


493-2 


t 




10 


V 


38.4 


4i-5 


44.6 


47-7 


50.7 


53-7 


56.7 


V 


10 




h 


1143.1 


1166.3 


1189.5 


1212 .7 


1236.0 


1259-3 


1282.5 


h 






t 


213.0 


263 .0 


3I3-0 


363-0 


4i3-o 


463.0 


5i3- 


t 




*5 


V 


26.27 


28.40 


30.46 


32.50 


34-53 


36.56 


38.58 


V 


15 




h 


1150.7 


1174.2 


1197.6 


1221 .0 


1244.4 


1267 .7 


1291 .1 


h 






t 


228.0 


278.0 


328.O 


378.0 


428.0 


478.0 


528.0 


t 




20 


V 


20.08 


21 .69 


23-25 


24.80 


26.33 


27-85 


29-37 


V 


20 




h 


1156.2 


1179.9 


1203.5 


1227. 1 


1250.6 


1274-1 


1297.6 


h 






t 


240.1 


290.1 


340.I 


390.I 


440.1 


490.1 


54o.i 


t 




25 


V 


16.30 


17 .60 


18.86 


20.10 


21.32 


22.55 


23-77 


V 


25 




h 


1160.4 


1 184. 4 


1208.2 


1231-9 


1255-6 


1279.2 


1302 .8 


h 






t 


250.4 


300.4 


350.4 


400.4 


45o.4 


500.4 


55o.4 


t 




30 


V 


13-74 


14-83 


15.89 


16.93 


17.97 


18.99 


20.00 


V 


30 




h 


1163.9 


1188.1 


1212 .1 


1236 .0 


12597 


1283.4 


1307-1 


h 






t 


259-3 


309-3 


359-3 


409-3 


459-3 


509-3 


559-3 


t 




35 


V 


11.89 


12.85 


13-75 


14.65 


15-54 


16.42 


17.30 


V 


35 




h 


1166.8 


1191.3 


1215-4 


1239.4 


1263.3 


1287. 1 


1310.8 


h 






t 


267.3 


317-3 


367.2 


417-3 


467-3 


517-3 


567-3 


t 




40 


V 


10.49 


n-33 


12.13 


12.93 


13.70 


14.48 


15-25 


V 


40 




h 


1169.4 


1 194.0 


1218.4 


1242 .4 


1266.4 


1290.3 


1314.1 


h 






t 


274-5 


324-5 


374-5 


424-5 


474-5 


524-5 


574-5 


t 




45 


V 


9-39 


10.14 


10.86 


n-57 


12 .27 


12 .96 


1365 


V 


45 




h 


1171 .6 


1196.6 


1221 .0 


1245.2 


1269.3 


1293.2 


1317.0 


h 






t 


281 .0 


33IO 


381.0 


431.0 


481 .0 


53i -o 


581.0 


t 




50 


V 


8.51 


9.19 


9.84 


10.48 


11 .11 


11.74 


12.36 


V 


5o 




h 


1173.6 


1198.8 


1223.4 


1247.7 


1271 .8 


1295.8 


I3I9-7. 


h 






t 


287.1 


337-1 


387-1 


437-1 


487.1 


537-1 


587.1 


t 




55 


V 


7.78 


8.40 


9.00 


9-59 


10.16 


10.73 


11.30 


V 


55 




h 


H75-4 


1 200 . 8 


1225.6 


1250.0 


1274.2 


1298. 1 


1322.0 


h 






t 


292 .7 


342.7 


392.7 


442.7 


492.7 


542.7 


592.7 


t 




60 


V 


7.17 


7-75 


8.30 


8.84 


9-36 


9.89 


10.41- 


V 


60 




h 


1177.0 


1202 .6 


1227.6 


1252. 1 


1276.4 


1300.4 


1324-3 


h 






t 


298.0 


348.0 


398 -o 


448.0 


498.0 


548.0 


598.0 


t 




65 


V 


6.65 


7.20 


7.70 


8.20 


8.69 


9.17 


9-65 


V 


65 




h 


1178.5 


1204.4 


1229.5 


1254.0 . 


1278.4 


1302.4 


1326.4 


h 






t 


302.9 


352.9 


402.9 


452.9 


502.9 


552.9 


602.9 


t 




70 


v. 


6.20 


6.71 


7.18 


7-65 


8. 11 


8.56 


9.01 


V 


70 




h 


1179.8 


1205.9 


1231 .2 


1255-8 


1280.2 


1304-3 


1328.3 


h 






t 


307.6 


357-6 


407.6 


457-6 


507.6 


557-6 


607.6 


t 




75 


V 


5.8i 


6.28 


6.73 


7.17 


7.60 


8.02 


8.44 


V 


75 




h 


1181.1 


1207.5 


1232 .8 


1257-5 


1282 .0 


1306. 1 


1330 - 1 


h 






t 


312.0 


362 .0 


412 .0 


462 .0 


512.0 


562.0 


612 .0 


t 




80 


V 


5-47 


5-92 


6-34 


6-75 


7.17 


7.56 


7-95 


V 


80 




h 


1182.3 


1208.8 


1234.3 


1259.0 


1283.6 


1307.8 


I33L9 


h 






t 


3i6-3 


366.3 


416.3 


466.3 


516.3 


566.3 


616.3 


t 




85 


V 


5-i6 


5-59 


6-99 


6.38 


6.76 


7.14 


7-51 


V 


85 




h 


1183.4 


1210. 2 


1235-8 


1260.6 


1285.2 


I309-4 


1333-5 


h 




t = 


Temperature, deg. Fahr. v = Specific volume, in cubic feet, per pou 


tid. 








h = Tot£ 


d heat froi 


n water at 


32 degree 


s, B.t.u. 









APPENDIX 

Table IV. — Continued 



203 



Pressure, 


Satu- 


Degrees of Superheat. 


Pressure, 


Pounds 
Absolute. 


rated 
Steam. 




Pounds 
Absolute. 




















50 


IOO 


150 


200 


250 


300 








t 


320.3 


370.3 


420.3 


470.3 


520.3 


570.3 


620.3 


t 




90 


V 


4.89 


5-29 


5-67 


6.04 


6.40 


6.76 


7 


11 


V 


90 




h 


1184.4 


I2II .4 


1237.2 


1262 .0 


1286.6 


1310.8 


1334 


9 


h 






t 


324-1 


374-1 


424.1 


474-1 


524-1 


574-1 


624 


1 


t 




95 


V 


4-65 


5-03 


5-39 


5-74 


6.09 


6-43 


6 


76 


V 


95 




h 


1185.4 


1212 .6 


1238.4 


1263.4 


1288. 1 


1312.3 


1336 


4 


h 






t 


327.8 


377-8 


427.8 


477-8 


527.8 


577-8 


627 


8 


t 




IOO 


V 


4-43 


4-79 


5-i4 


5-47 


5.80 


6.12 


6 


44 


V 


IOO 




h 


1186.3 


1213.8 


1239.7 


1264.7 


1289.4 


1313-6 


1337 


8 


h 






t 


331-4 


381.4 


431-4 


481.4 


531-4 


581.4 


631 


4 


t 




105 


V 


4-23 


4-58 


4.91 


5-23 


5-54 


5-8 5 


6 


15 


V 


105 




h 


1187.2 


1214.9 


1 240 . 8 


1265.9 


1290.6 


13*4-9 


1339 


1 


h 






t 


334-8 


384-8 


434-8 


484.8 


534-8 


584.8 


634 


8 


t 




no 


V 


4-05 


4-38 


4.70 


5-oi 


5-3i 


5-6i 


5 


90 


V 


no 




h 


1188.0 


1215.9 


1242 .0 


1267. 1 


1291 .9 


1316.2 


1340 


4 


h 






t 


338.1 


388.1 


438.1 


488.1 


538.1 


588.1 


638 


1 


t 




115 


V 


3-88 


4.20 


4-5i 


4.81 


5-09 


5-38 


5 


66 


V 


II 5 




h 


1188.8 


1216.9 


1243. 1 


1268.2 


1293.0 


I3I7-3 


1341 


5 


h 






t 


341-3 


391-3 


441-3 


491-3 


541-3 


591-3 


641 


3 


t 




120 


V 


3-73 


4.04 


4-33 


4.62 


4.89 


5-i7 


5 


44 


V 


120 




h 


1189.6 


1217.9 


1244. 1 


1269.3 


1 294. 1 


1318.4 


1342 


7 


h 






t 


' 344-4 


394-4 


444.4 


494.4 


544-4 


594-4 


644 


4 


t 




125 


V 


3-58 


3.88 


4-i7 


4-45 


4-7i 


4-97 


5 


23 


V 


I2 5 




h 


1190.3 


1218.8 


1245. 1 


1270.4 


1295.2 


I3I9-5 


1343 


8 


h 






t 


347-4 


397-4 


447-4 


497-4 


547-4 


597-4 


647 


4 


t 




130 


V 


3-45 


3-74 


4.02 


4.28 


4-54 


4.80 


5 


05 


V 


130 




h 


1191 .0 


1219.7 


1246. 1 


1271 .4 


1296.2 


1320.6 


1344 


9 


h 






t 


35o.3 


400.3 


45o.3 


500.3 


55o.3 


600.3 


650 


3 


t 




135 


V 


3-33 


3.61 


3.88 


4.14 


4.38 


4-63 


4 


87 


V 


J 35 




h 


1191 .6 


1220.6 


1247.0 


1272.3 


1297 .2 


1321 .6 


1345 


9 


h 






t 


353 - 1 


403- 1 


453-1 


503- 1 


553-1 


603.1 


653 


1 


t 




140 


V 


3-22 


3-49 


3-75 


4.00 


4.24 


4.48 


4 


7i 


V 


140 




h 


1192 .2 


1221 .4 


1248.0 


1273-3 


1298.2 


1322 .6 


1346 


9 


h 






t 


355-8 


405.8 


455-8 


505-8 


555-8 


605.8 


655 


8 


t 




145 


V 


3.12 


3-38 


3 63 


3-87 


4.10 


4-33 


4 


56 


V 


J 45 




h 


1192 .8 


1222 .2 


1248.8 


1274.2 


1 299. 1 


1323.6 


1347 


9 


h 






t 


358.5 


408.5 


458.5 


508.5 


558.5" 


608.5 


658 


5 


t 




150 


V 


3.01 


3-27 


3-5i 


3-75 


3-97 


4.19 


4 


41 


V 


150 




h 


H93-4 


1223 .0 


1249.6 


1275. 1 


1300.0 


1324-5 


1348 


8 


h 






t 


361.0 


411 .0 


461 .0 


511. 


561.0 


611 .0 


661 





t 




155 


V 


2 .92 


3-*7 


3-4i 


3-t>3 


3-85 


4.06 


4 


28 


V 


!55 




h 


1 194.0 


1223 .6 


1250.5 


1276.0 


1300.8 


13253 


1349 


7 


h 






t 


363-6 


413.6 


463-6 


513-6 


563-6 


613.6 


663 


6 


t 




160 


V 


2.83 


3-07 


3-3o 


3-53 


3-74 


3-95 


4 


15 


V 


160 




h 


"94-5 


1224.5 


1251.3 


1276.8 


1301.7 


1326 . 2 


1350 


6 


h 






t 


366.0 


416.0 


466.0 


516.0 


566.0 


616.0 


666 





t 




165 


V 


2-75 


2.99 


3.21 


3-43 


3 64 


3 84 


4 


04 


V 


165 




h 


1195.0 


1225 . 2 


1252.0 


1277.6 


1302.5 


1327-1 


i35i 


5 


h 






t 


368.5 


418.5 


468.5 


518.5 


568.5 


618.5 


668 


5 


t 




170 


V 


2.68 


2 .91 


3.12 


3-34 


3-54 


3-73 


3 


92 


V 


170 




h 


II95-4 


1225.9 


1252.8 


1278.4 


I303-3 


1327.9 


1352.3 


h 





t — Temperature, deg. Fahr. v = Specific volume, in cubic feet, per pound. 

h = Total heat from water at 32 degrees, B.t.u. 



204 



APPENDIX 

Table IV. — Continued 



Pressure, 


Satu- 


Degrees of Superheat. 


Pressure. 


Pounds 






T~% 




Absolute. 


IdLLU 

Steam. 














irounas 
Absolute 








So 


100 


150 


200 


250 


300 








t 


370.8 


420.8 


470.8 


520.8 


57o.8 


620.8 


670.8 


t 




175 


V 


2 .60 


2.83 


3 04 


3-24 


3-44 


3 63 


3-82 


V 


175 




h 


H95-9 


1226.6 


1253.6 


1279. 1 


1304-1 


1328.7 


1353-2 


h 






t 


373-1 


423.1 


473-1 


523-1 


573-1 


623.1 


673.1 


t 




180 


V 


2-53 


2-75 


2 .96 


3.16 


3-35 


3-54 


3-72 


V 


180 




h 


1196.4 


1227. 2 


1254-3 


1279.9 


1304.8 


13295 


1353-9 


h 






t 


375-4 


425-4 


475-4 


525-4 


575-4 


625.4 


675-4 


t 




185 


V 


2.47 


2.68 


2.89 


3.08 


3-27 


3-45 


3 63 


V 


185 




h 


1196.8 


1227.9 


1255-0 


1280.6 


1305-6 


1330.2 


1354-7 


h 






t 


377-6 


427.6 


477.6 


527.6 


577-6 


627.6 


677.6 


t 




190 


V 


2 .41 


2 .62 


2.81 


3.00 


3-19 


3-37 


3-55 


V 


190 




h 


H97-3 


1228.6 


1255-7 


1281.3 


1306.3 


1330.9 


1355-5 


h 






t 


379-8 


429.8 


479.8 


529.8 


579-8 


629.8 


679.8 


t 




195 


V 


2-35 


2-55 


-2.75 


2-93 


3-ii 


3-29 


3 46 


V 


i95 




h 


1197.7 


1229.2 


1256.4 


1282 .0 


1307.0 


1331-6 


1356.2 


h 






t 


381.9 


431-9 


481.9 


531-9 


581.9 


631.9 


681.9 


t 




200 


V 


2 .29 


2.49 


2.68 


2.86 


3-04 


3.21 


3-38 


V 


200 




h 


1198.1 


1229.8 


1257. 1 


1282 .6 


1307.7 


1332.4 


1357 


h 






t 


384.0 


434 -o 


484.0 


534 


584.0 


634.0 


684.0 


t 




205 


V 


2.24 


2-44 


2 .62 


2.80 


2.97 


3 14 


3 30 


V 


205 




h 


1198.5 


1230.4 


1257-7 


1283.3 


1308.3 


i333-o 


1357-7 


h 






t 


386.0 


436.0 


486.0 


536.o 


586.0 


636.0 


686.0 


t 




210 


V 


2 .19 


2.38 


2.56 


2.74 


2 .91 


3-07 


3-23 


V 


210 




h 


1198.8 


1 231 .0 


1258.4 


1284.0 


1309.0 


1333-7 


1358.4 


h 






t 


388.0 


438.0 


488.0 


538.o 


588.0 


638.0 


688.0 


t 




215 


V 


2 .14 


2-33 


2.51 


2.68 


2.84 


3.00 


3.16 


V 


215 




h 


1199.2 


1231 .6 


1259.0 


1284.6 


1309 . 7 


1334-4 


1359- 1 


h 






t 


389-9 


439-9 


489.9 


539-9 


589-9 


639-9 


689.9 


t 




220 


V 


2 .09 


2.28 


2-45 


2 .62 


2.78 


2-94 


3.10 


V 


220 




h 


1199.6 


1232 .2 


1259.6 


1285 .2 


i3 IO -3 


1335- 1 


1359-8 


h 






t 


39 J -9 


441.9 


491.9 


541-9 


591-9 


641.9 


691.9 


t 




225 


V 


2.05 


2.23 


2 .40 


2-57 


2 .72 


2.88 


3-03 


V 


225 




h 


1199.9 


1232.7 


1260.2 


1285.9 


1310.9 


1335-7 


1360.3 


h 






t 


393-8 


443-8 


493-8 


543-8 


593-8 


643-8 


693-8 


t 




230 


V 


2 .00 


2.18 


2-35 


. 2.51 


2 .67 


2.82 


2.97 


V 


230 




h 


1200.2 


1233.2 


1260.7 


1286.5 


1311.6 


1336.3 


1361 .0 


h 






t 


395-6 


445-6 


495-6 


545-6 


595-6 


645.6 


695.6 


t 




235 


V 


1 .96 


2 .14 


2.30 


2.46 


2 .62 


2-77 


2 .91 


V 


235 




h 


1 200 . 6 


1233-8 


1261 .4 


1287. 1 


1312 . 2 


I337-0 


1361.7 


h 






t 


397-4 


447-4 


497-4 


547-4 


597-4 


647.4 


697.4 


t 




240 


V 


1 .92 


2 .09 


2 .26 


2 .42 


2-57 


2 .71 


2.85 


V 


240 




h 


1200.9 


1234-3 


1261 .9 


1287.6 


1312.8 


1337-6 


1362.3 


h 






t 


399-3 


449-3 


499-3 


549-3 


599-3 


649-3 


699-3 


t 




245 


V 


1.89 


2.05 


2 .22 


2-37 


2.52 


2.66 


2.80 


V 


245 




h 


1 201 .2 


1234.8 


1262 .5 


1288.2 


I3I3-3 


1338.2 


1362.9 


h 






t 


401 .0 


451.0 


501.0 


55i -o 


601 .0 


651.0 


701 .0 


t 




250 


V 


1-85 


2 .02 


2 .17 


2-33 


2-47 


2 .61 


2-75 


V 


250 




h 


1201.5 


1235-4 


1263 .0 


1288.8 


13*3-9 


1338.8 


1363-5 


h 






t 


402 .8 


452.8 


502.8 


552.8 


602.8 


652.8 


702.8 


t 




255 


V 


1. 81 


1.98 


2 .14 


2.28 


2-43 


2.56 


2 .70 


V 


255 




h 


1 201 .8 


12359 


1263 .6 


1289.3 


1314-5 


1339-3 


1364-1 


h 





t = 



Temperature, deg. Fahr. v = Specific volume, in cubic feet, per pound. 

h = Total heat from water at 32 degrees, B.t.u. 



APPENDIX 



205 



Table V. — SATURATED VAPOR OF SULPHUR DIOXIDE 

Reproduced by permission from Peabody's " Steam and Entropy Tables." 

English Units. 



fii 


uA 


<u 




G 


-»-> 
g 


-4-> 

G 
0>«-i 




a 


Temperatui 

Degrees Fa 

renheit. 


Pressure, 
Pounds pe 
Square Inc 


•G . 

"o'g 
-n cr ■ 

<ut-3 


■w 


Q> U 
l-H O 
HH ft 

> 


Heat Equiva 
of Interna 
Work. 


Heat Equival 
of Externa 
Work. 


P. o* 

c 


!3 

> 


'0 
a> 
Pi 


t 


P 


A 


H 


L 


P 


Apu 


9 


5 


-40 


3-14 


-29 


166 


195 


182 


13 


—0.0632 


23.O 


— -20 


5 90 


— 21 


169 


190 


176 


14 


-0.0447 


12.7 


O 


io.35 


-13 


172 


185 


170 


15 


—0.0268 


7-54 


IO 


i3-4i 


- 9 


173 


182 


167 


15 


—0.0182 


5-93 


20 


I7-I5 


- 5 


174 


179 


164 


15 


—0.0098 


4.72 


30 


21 .66 


— I 


176 


177 


162 


15 


—0.0016 


3.8i 


40 


27.06 


3 


177 


174 


158 


16 


. 0064 


3.10 


50 


33-45 


7 


178 


171 


155 


16 


0.0144 


2.58 


60 


40.98 


11 


179 


168 


152 


16 


0.0221 


2. 11 


70 


49-75 


15 


181 


166 


ISO 


16 


0.0297 


1.78 



Table VI. — PROPERTIES OF CARBON DIOXIDE 

Reproduced from Marks' " Mechanical Engineers' Handbook." 





.fe-8 


Heat Content 


G 


UO G 
u O 


G 

13 13 




S 
3 


"o 


-nperatu 
grecs F, 
renheit 


J3 coi-i 






13 n 




; Equiv 
Extern 
Work 


*o 

> 


§•3 


w G £; 
<ij 3 ra 

^ O 3 


of 


of 


u 


*c3 

G P< 
u c3 





u a 1 


53^ 




Liquid. 


Vapor. 


> 


G-— 


3 ° 




(D 
Pi 


m 












►h O 


03 




t 


£ 


h ' 


H 


z. 


P 


.A^M 


s 





20 


221 .O 


-24-75 


IOO.50 


125-25 


IO9.O 


16.3 


0.4173 


-O.0513 





308.O 


— 16.00 


101 .00 


117 


00 


IOI 


3 


15-7 


0.2918 


-O.0325 


10 


362.5 


-11.36 


100.89 


112 


35 


96 


9 


15-4 


0.2450 


— O.0227 


20 


421 .6 


— 6.40 


100.43 


I06 


83 


91 


8 


150 


. 2060 


— O.OI26 


30 


488.8 


— 1 .04 


99-43 


IOO 


47 


86 


7 


14.2 


0.1724 


— 0.002I 


40 


564-5 


4-36 


98.25 


93 


89 


80 


4 


13-5 


0.1444 


O.O087 


50 


650.0 


10.76 


96.30 


85 


54 


73 


3i 


12.2 


0.1205 


O.O205 


60 


744.o 


17-85 


93-54 


75 


69 


64 


90 


10.8 


0.0986 


O.0334 


70 


847.0 


26.02 


89-35 


63 


33 


54 


03 


9-3 


0.0816 


O.0483 



Table VII. — NAPERIAN LOGARITHMS 

Reproduced by permission from Goodenough's " Properties of Steam and Ammonia." 

e =2.7182818 log e = 0.4342945 = M 



1.0 

1.1 
1.2 
1.3 

1.4 
1.5 
1.6 

1.7 
1.8 
1.9 

2.0 

2.1 
2.2 
2.3 

2.4 
2.5 
2.6 

2.7 
2.8 
2.9 

3.0 

3.1 
3.2 
3.3 

3.4 
3.5 
3.6 

3.7 
3.8 
3.9 

4.0 

4.1 
4.2 
4.3 

4.4 
4.5 
4.6 

4.7 
4.8 
4.9 

5.0 

5.1 
5.2 
5.3 

5.4 
5.5 
5.6 



0. 0000 

0.09531 

0.1823 

0.2624 

0.3365 
0. 4055 
0. 4700 

0.5306 
0.5878 
0.6418 

0.6931 

0.7419 
0.7884 
0.8329 

0.8755 
0.9163 
0.9555 

0.9933 
1.0296 
1.0647 

1.0986 

1.1314 
1.1632 
1.1939 

1.2238 
1.2528 
1.2809 

1.3083 
1.3350 
1.3610 

1.3863 

1.4110 
1.4351 
1.4586 

1.4816 
1.5041 
1.5261 

1.5476 

1.5686 
1.5892 

1.6094 

1.6292 
1.6487 
1.6677 

1.6864 
1.7047 
1.7228 



1.6114 

1.6312 
1.6506 
1.6696 

1.6882 
1.7066 
1.7246 



0.00995 

0. 1044 
0. 1906 
0. 2700 

0.3436 
0.4121 
0.4762 

0.5365 
0.5933 
0.6471 

0.6981 

0.7467 
0.7930 
0.8372 

0.8796 
0.9203 
0.9594 

0.9969 

1.0332 

0682 

1.1019 

1.1346 
1.1663 
1.1969 

1.2267 
1.2556 
1.2837 

1.3110 
1.3376 
1.3635 

1.3888 

1.4134 
1.4375 
1.4609 

1.4839 
1.5063 
1.5282 

1.5497 
1.5707 
1.5913 



0.01980 

0.1133 
0. 1988 
0.2776 

0.3507 
0.4187 
0. 4824 

0.5423 
0.5988 
0.6523 

0.7031 

0.7514 
0.7975 
0.8416 

0.8838 
0.9243 
0.9632 

1.0006 
1.0367 
1.0716 

1.1053 



1.1378 1.1410 
1.1694 1.1725 
1.2000 1.2030 



0.02956 

0.1222 
0.2070 
0. 2852 

0.3577 
0.4253 
0. 4886 

0.5481 
0. 6043 
0.6575 

0. 7080 

0.7561 
0.8020 
0.8459 

0.8879 
0.9282 
0.9670 

1.0043 
1.0403 
1.0750 



1.1086 1.1119 



0.03922 

0.1310 
0.2151 
0.2927 

0.3646 
0.4318 
0.4947 

0.5539 
0.6098 
0.6627 

0.7129 

0.7608 
0.8065 
0. 8502 

0. 8920 

0.9322 
0.9708 

1.0080 
1.0438 
1.0784 



1.2296 
1.2585 
1.2865 

1.3137 
1.3403 
1.3661 

1.3913 

1.4159 
1.4398 
1.4633 

1.4861 
1.5085 
1.5304 

1.5518 
1.5728 
1.5933 

1.6134 



1.6332 
1.6525 
1.6715 

1.6901 
1.7884 
1.7263 



1.2326 
1.2613 
1.2892 

1.3164 
1.3429 
1.3686 

1.3938 

1.4183 
1.4422 
1.4656 

1.4884 
1.5107 
1.5326 

1.5539 
1.5748 
1.5953 

1.6154 

1.6351 
1.6544 
1.6734 

1.6919 
1.7102 
1.7281 



1.1442 
1.1756 
1.2060 

1.2355 
1.2641 
1.2920 

1.3191 
1.3455 
1.3712 

1.3962 

1.4207 
1.4446 
1.4679 

1.4907 
1.5129 
1.5347 

1.5560 
1.5769 
1.5974 

1.6174 

1.6371 
1.6563 
1.6752 

1.6938 
1.7120 
1.7299 



206 



0. 04879 

0. 1398 
0.2231 
0.3001 

0.3716 
0.4382 
0.5008 

0.5596 
0.6152 
0. 6678 

0.7178 

0.7655 
0.8109 
0. 8544 

0.8961 
0.9361 
0.9746 

1.0116 
1.0473 
1.0818 

1.1151 

1.1474 
1.1787 
1.2090 

1.2384 
1.2669 
1.2947 

1.3218 
1.3481 
3737 

1.3987 

1.4231 
1.4469 
1.4702 

1.4929 
1.5151 
1.5369 



1.5581 
1.5790 
1.5994 

1.6194 

1.6390 
1.6582 
1.6771 

1.6956 
1.7138 
1.7317 



0.05827 

0. 1484 
0.2311 
0.3075 

0.3784 
0.4447 
0.5068 

0. 5653 
0. 6206 
0.6729 

0. 7227 

0.7701 
0.8154 
0. 8587 

0.9002 
0.9400 
0.9783 

1.0152 
1.0508 
1.0852 

1.1184 

1.1506 
1.1817 
1.2119 

1.2413 
1.2698 
1.2975 

1.3244 
1.3507 
1.3762 



0. 06766 

0. 1570 
0.2390 
0.3148 

0.3853 
0.4511 
0.5128 

0.5710 
0.6259 
0. 6780 

0.7275 

0. 7747 
0.8198 
0.8629 

0.9042 
0.9439 
0.9821 

1.0188 
1.0543 
1.0886 

1.1217 

1.1537 
1.1848 
1.2149 

1.2442 
1.2726 
1.3002 

1.3271 
1.3533 

1.3788 



8 



1.4012 1.4036 1.4061 



1.4255 
1.4493 
1.4725 

1.4951 
1.5173 
1.5390 



1.5602 
1.5810 
1.6014 

1.6214 

1.6409 
1.6601 
1.6790 

1.6974 
1.7156 
1.7334 



0. 07696 

0. 1655 
0.2469 
0. 3221 

0.3920 
0.4574 
0.5188 

0. 5766 
0.6313 
0.6831 

0.7324 

0.7793 
0.8242 
0.8671 

0.9083 
0. 9478 
0. 9858 



0.08618 

0. 1739 
0.2546 
0.3293 

0.3988 
0.4637 
0.5247 

0. 5822 
0.6366 
0.6881 

0.7372 

0. 7839 
0.8286 
0.8713 

0.9123 

0.9517 
0.9895 



1.0225 1.0260 

1.0578 1.0613 

1.0919 1.0953 

1.1249 1.1282 



1.1569 

1.1878 
1.2179 

1.2470 
1.2754 
1.3029 

1.3297 
1.3558 
1.3813 



1.4279 
1.4516 
1.4748 

1.4974 
1.5195 
1.5412 

1.5623 
1.5831 
1.6034 

1.6233 

1.6429 
1.6620 
1.6808 

1.6993 
1.7174 
1.7352 



1.4303 
1.4540 
1.4770 

1.4996 
1.5217 
1.5433 

1.5644 
1.5851 
1.6054 

1.6253 

1.6448 
1.6639 
1.6827 

1.7011 
1.7192 
1.7370 



1.1600 
1.1909 
1.2208 



1.2499 
1.2782 
1.3056 

1.3324 
1.3584 
1.3838 

1.4085 

1.4327 
1.4563 
1.4793 

1.5019 
1.5239 
1.5454 

1.5665 
1.5872 
1.6074 

1.6273 

1.6467 
1.6658 
1.6845 

1.7029 
1.7210 
1.7387 



Table VII. — Continued. NAPERIAN LOGARITHMS 








1 


2 


3 


4 


5 


6 


7 


8 


9 


5.7 
5.8 
5.9 


1.7405 
1.7579 
1.7750 


1.7422 
1.7596 
1.7766 


1.7440 
1.7613 
1.7783 


1.7457 
1.7630 
1.7800 


1.7475 
1.7647 
1.7817 


1.7492 
1.7664 
1.7834 


1.7509 
1.7681 
1.7851 


1.7527 
1.7699 
1.7867 


1.7544 
1.7716 
1.7884 


1.7561 
1.7733 
1.7901 


6.0 


1.7918 


1.7934 


1.7951 


1.7967 


1.7984 


1.8001 


1.8017 


1.8034 


1.8050 


1.8066 


6.1 
6.2 
6.3 


1.8083 
1.8245 
1.8405 


1.8099 
1.8262 
1.8421 


1.8116 
1.8278 
1.8437 


1.8132 
1.8294 
1.8453 


1.8148 
1.8310 
1.8469 


1.8165 
1.8326 
1.8485 


1.8181 
1.8342 
1.8500 


1.8197 
1.8358 
1.8516 


1.8213 
1.8374 
1.8532 


1.8229 
1.8390 
1.8547 


6.4 
6.5 
6.6 


1.8563 
1.8718 
1.8871 


1.8579 
1.8733 
1.8886 


1.8594 
1.8749 
1.8901 


1.8610 

1.8764 
1.8916 


1.8625 
1.8779 
1.8931 


1.8641 
1.8795- 
1.8946 


1.8656 
1.8810 
1.8961 


1.8672 
1.8825 
1.8976 


1.8687 
1.8840 
1.8991 


1.8703 

1.8856 
1.9006 


6.7 
6.8 
6.9 


1.9021 
1.9169 
1.9315 


1.9036 
1.9184 
1.9330 


1.9051 
1.9199 
1.9344 


1.9066 
1.9213 
1.9359 


1.9081 
1.9228 
1.9373 


1.9095 
1.9242 
1.9387 


1.9110 
1.9257 
1.9402 


1.9125 
1.9272 
1.9416 


1.9140 
1.9286 
1.9430 


1.9155 
1.9301 
1.9445 


7.0 


1.9459 


1.9473 


1.9488 


1.9502 


1.9516 


1.9530 


1.9544 


1.9559 


1.9573 


1.9587 


7.1 
7.2 
7.3 


L9601 
1.9741 
1.9879 


1.9615 
1.9755 
1.9892 


1.9629 
1.9769 
1.9906 


1.9643 

1.9782 
1.9920 


1.9657 
1.9796 
1.9933 


1.9671 
1.9810 
1.9947 


1.9685 
1.9824 
1.9961 


1.9699 
1.9838 
1.9974 


1.9713 

1.9851 
1.9988 


1.9727 
1.9865 
2.0001 


7.4 
7.5 
7.6 


2.0015 
2.0149 
2.0281 


2.0028 
2.0162 
2.0295 


2. 0042 
2.0176 
2.0308 


2.0055 
2.0189 
2.0321 


2.0069 
2.0202 
2. 0334 


2. 0082 
2.0215 
2. 0347 


2.0096 
2.0229 
2.0360 


2.0109 
2.0242 
2.0373 


2.0122 
2.0255 
2.0386 


2.0136 

2.0268 
2.0399 


7.7 
7.8 
7.9 


2.0412 
2.0541 
2.0668 


2. 0425 
2.0554 
2.0681 


2.0438 
2.0567 
2.0694 


2.0451 
2.0580 
2.0707 


2. 0464 
2.0592 
2.0719 


2. 0477 
2.0605 
2.0732 


2. 0490 
2.0618 
2.0744 


2.0503 
2.0631 
2.0757 


2.0516 
2.0643 
2.0769 


2.0528 
2.0656 
2. 0782 


8.0 


2.0794 


2.0807 


2.0819 


2.0832 


2.0844 


2.0857 


2.0869 


2.0881 


2.0894 


2.0906 


8.1 
8.2 
8.3 


2.0919 
2.1041 
2-1163 


2.0931 
2.1054 
2.1175 


2.0943 
2.1066 
2.1187 


2.0956 
2.1078 
2.1199 


2.0968 
2.1090 
2.1211 


2. 0980 
2.1102 
2. 1223 


2.0992 
2.1114 
2.1235 


2. 1005 
2.1126 
2. 1247 


2.1017 
2.1138 
2.1258 


2. 1029 
2.1150 
2. 1270 


8.4 
8.5 
8.6 


2. 1282 
2.1401 
2.1518 


2. 1294 
2.1412 
2.1529 


2.1306 
2.1424 
2. 1541 


2.1318 
2.1436 
2.1552 


2.1330 
2. 1448 
2.1564 


2.1342 
2.1459 
2.1576 


2.1353 
2.1471 
2.1587 


2.1365 
2. 1483 
2. 1599 


2. 1377 
2.1494 
2.1610 


2.1389 
2.1506 
2. 1622 


8.7 
8.8 
8.9 


2.1633 
2. 1748 
2.1861 


2. 1645 
2.1759 
2.1872 


2.1656 
2. 1770 
2. 1883 


2.1668 
2.1782 
2. 1894 


2.1679 
2.1793 
2. 1905 


2.1691 
2.1804 
2.1917 


2. 1702 
2.1815 
2. 1928 


2.1713 

2.1827 
2.1939 


2.1725 

2.1838 
2.1950 


2.1736 
2.1849 
2.1961 


9.0 


2.1972 


2.1983 


2.1994 


2.2006 


2.2017 


2.2028 


2.2039 


2.2050 


2.2061 


2.2072 


9.1 
9.2 
9.3 


2.2083 
2.2192 
2.2300 


2.2094 
2.2203 
2.2311 


2.2105 

2.2214 
2.2322 


2.2116 
2.2225 
2.2332 


2.2127 
2.2235 
2.2343 


2.2138 
2. 2246 
2.2354 


2.2148 
2.2257 
2.2364 


2.2159 
2.2268 
2.2375 


2.2170 

2.2279 
2.2386 


2.2181 
2.2289 
2.2396 


9.4 
9.5 
9.6 


2. 2407 
2.2513 
2.2618 


2.2418 
2.2523 
2.2628 


2. 2428 
2.2534 
2.2638 


2. 2439 
2.2544 
2.2649 


2. 2450 
2. 2555 
2.2659 


2.2460 
2.2565 
2. 2670 


2.2471 
2.2576 
2.2680 


2.2481 
2.2586 
2.2690 


2.2492 
2.2597 
2.2701 


2. 2502 
2.2607 
2.2711 


9.7 
9.8 
9.9 


2.2721 

2.2824 
2. 2925 


2.2732 
2.2834 
2.2935 


2.2742 
2. 2844 
2.2946 


2.2752 
2.2854 
2.2956 


2.2762 
2.2865 
2.2966 


2.2773 
2.2875 
2.2976 


2.2783 
2. 2885 
2.2986 


2.2793 
2.2895 
2.2996 


2.2803 
2.2905 
2.3006 


2.2814 
2.2915 
2.3016 


10.0 


2.3026 





















207 



Table VIII. — LOGARITHMS 

Reproduced by permission from Goodenough's " Properties of Steam and Ammonia." 



Nat. 
Nos. 






















Proportional Parts. 





1 


2 


3 


4 


5 


6 


7 


8 


9 






































1 2 


3 


4 


5 


6 


7 


8 9 


10 


0000 


0043 


0086 


0128 


0170 


0212 


0253 


0294 


0334 


0374 


4 8 


12 


17 21 


25 


29 


33 37 


11 


0414 


0453 


0492 


0531 


0569 


0607 


0645 


0682 


0719 


0755 


4 8 


11 


15 


19 


23 


26 


30 34 


12 


0792 


0828 


0864 


0899 


0934 


0969 


1004 


1038 


1072 


1106 


3 7 


10 


14 


17 


21 


24 


28 31 


13 


1139 


1173 


1206 


1239 


1271 


1303 


1335 


1367 


1399 


1430 


3 6 


10 


13 


16 


19 


23 


26 29 


14 


1461 


1492 


1523 


1553 


1584 


1614 


1644 


1673 


1703 


1732 


3 6 


9 


12 


15 


18 


21 


24 27 


15 


1761 


1790 


1818 


1847 


1875 


1903 


1931 


1959 


1987 


2014 


3 6 


8 


11 


14 


17 


20 


22 25 


16 


2041 


2068 


2095 


2122 


2148 


2175 


2201 


2227 


2253 


2279 


3 5 


8 


11 


13 


16 


18 


21 24 


17 


2304 


2330 


2355 


2380 


2405 


2430 


2455 


2480 


2504 


2529 


2 5 


7 


10 


12 


15 


17 


20 22 


18 


2553 


2577 


2601 


%25 


2648 


2672 


2695 


2718 


2742 


2765 


2 5 


7 


9 


12 


14 


16 


19 21 


19 


2788 


2810 


2833 


2856 


2878 


2900 


2923 


2945 


2967 


2989 


2 4 


7 


9 


11 


13 


16 


18 20 


20 


3010 


3032 


3054 


3075 


3096 


3118 


3139 


3160 


3181 


3201 


2 4 


6 


8 


11 


13 


15 


17 19 


21 


3222 


3243 


3263 


3284 


3304 


3324 


3345 


3365 


3385 


3404 


2 4 


6 


8 


10 


12 


14 


16 18 


22 ' 


3424 


3444 


3464 


3483 


3502 


3522 


3541 


3560 


3579 


3598 


2 4 


6 


8 


10 


12 


14 


15 17 


23 


3617 


3636 


3655 


3674 


3692 


3711 


3729 


3747 


3766 


3784 


2 4 


6 


7 


9 


11 


13 


15 17 


24 


3802 


3820 


3838 


3856 


3874 


3892 


3909 


3927 


3945 


3962 


2 4 


5 


7 


9 


11 


12 


14 16 


25 


3979 


3997 


4014 


4031 


4048 


4065 


4082 


4099 


4116 


4133 


2 3 


5 


7 


9 


10 


12 


14 15 


26 


4150 


4166 4183 


4200 


4216 


4232 


4249 


4265 


4281 


4298 


2 3 


5 


7 


8 


10 


11 


13 15 


27 


4314 


4330 


4346 


4362 


4378 


4393 


4409 


4425 


4440 


4456 


2 3 


5 


6 


8 


9 


11 


13 14 


28 


4472 


4487 


4502 


4518 


4533 


4548 


4564 


4579 


4594 


4609 


2 3 


5 


6 


8 


9 


11 


12 14 


29 


4624 


4639 


4654 


4669 


4683 


4698 


4713 


4728 


4742 


4757 


1 3 


4 


6 


7 


9 


10 


12 13 


30 


4771 


4786 


4800 


4814 


4829 


4843 


4857 


4871 


4886 


4900 


1 3 


4 


6 


7 


9 


10 


11 13 


31 


4914 


4928 


4942 


4955 


4969 


4983 


4997 


5011 


5024 


5038 


1 3 


4 


6 


7 


8 


10 


11 12 


32 


5051 


5065 


5079 


5092 


5105 


5119 


5132 


5145 


5159 


5172 


1 3 


4 


5 


7 


8 


9 


11 12 


33 


5185 


5198 


5211 


5224 


5237 


5250 


5263 


5276 


5289 


5302 


1 3 


4 


5 


6 


8 


9 


10 12 


34 


5315 


5328 


5340 


5353 


5366 


5378 


5391 


5403 


5416 


5428 


1 3 


4 


5 


6 


8 


9 


10 11 


35 


5441 


5453 


5465 


5478 


5490 5502 


5514 


5527 


5539 


5551 


1 2 


4 


5 


6 


7 


9 


10 11 


36 


5563 


5575 


5587 


5599 


56115623 


5635 


5647 


5658 


5670 


1 2 


4 


5 


6 


7 


8 


10 11 


37 


5682 


5694 


5705 


5717 


5729 5740 


5752 


5763 


5775 


5786 


1 2 


3 


5 


6 


7 


8 


9 10 


38 


5798 


5809 


5821 


5832 


5843 


5855 


5866 


5877 


5888 


5899 


12 


3 


5 


6 


7 


8 


9 10 


39 


5911 


5922 


5933 


5944 


5955 


5966 


5977 


5988 


5999 


6010 


1 2 


3 


4 


5 


7 


8 


9 10 


40 


6021 


6031 


6042 


6053 


6064 


6075 


6085 


6096 


6107 


6117 


1 2 


3 


4 


5 


6 


8 


9 10 


41 


6128 


6138 


6149 


6160 


6170 


6180 


6191 


6201 


6212 


6222 


1 2 


3 


4 


5 


6 


7 


8 9 


42 


6232 


6243 


6253 


6263 


6274 


6284 


6294 


6304 


6314 


6325 


1 2 


3 


4 


5 


6 


7 


8 9 


43 


6335 


6345 


6355 


6365 


6375 


6385 


6395 


6405 


6415 


6425 


1 2 


3 


4 


5 


6 


7 


8 9 


44 


6435 


6444 


6454 


6464 


6474 


6484 


6493 


6503 


6513 


6522 


1 2 


3 


4 


5 


6 


7 


8 9 


45 


6532 


6542 


6551 


6561 


6571 


6580 


6590 


6599 


6609 


6618 


1 2 


3 


4 


5 


6 


7 


8 9 


46 


6628 


6637 


6646 


6656 


6665 


6675 


6684 


6693 


6702 


6712 


1 2 


3 


4 


5 


6 


7 


7 8 


47 


6721 


6730 


6739 


6749 


6758 


6767 


6776 


6785 


6794 


6803 


1 2 


3 


4 


5 


5 


6 


7 8 


48 


6812 


6821 


6830 


6839 


6848 


6857 


6866 


6875 


6884 


6893 


1 2 


3 


4 


4 


5 


6 


7 8 


49 


6902 


6911 


6920 


6928 


6937 


6946 


6955 


6964 


6972 


6981 


1 2 


3 


4 


4 


5 


6 


7 8 


50 


6990 


6998 


7007 


7016 


7024 


7033 


7042 


7050 


7059 


7067 


1 2 


3 


3 


4 


5 


6 


7 8 


51 


7076 


7084 


7093 


7101 


7110 


7118 


7126 


7135 


7143 


7152 


1 2 


3 


3 


4 


5 


6 


7 8 


52 


7160 


7168 


7177 7185 


7193 


7202 


7210 


7218 


7226 


7235 


1 2 


2 


3 


4 


5 


6 


7 7 


53 


7243 


7251 


7259 


7267 


7275 


7284 


7292 


7300 


7308 


7316 


1 2 


2 


3 


4 


5 


6 


6 7 


54 


7324 


7332 


7340 


7348 


7356 


7364 


7372 


7380 


7388 


7396 


1 2 


2 


3 


4 


5 


6 


6 7 



208 



Table VIII. — Continued. LOGARITHMS 

























Proportional Parts. 


Nat. 
Nos. 





1 


2 


3 


4 


5 


6 


7 


8 


9 
















































12 3 


4 


5 


6 


7 


8 9 


55 


7404 


7412 


7419 


7427 


7435 


7443 


7451 


7459 


7466 


7474 


1 2 2 


3 


4 


5 


5 


6 7 


56 


7482 


7490 


7497 


7505 


7513 


7520 


7528 


7536 


7543 


7551 


12 2 


3 


4 


5 


5 


6 7 


57 


7559 


7566 


7574 


7582 


7589 


7597 


7604 


7612 


7619 


7627 


1 2 2 


3 


4 


5 


5 


6 7 


58 


7634 


7642 


7649 


7657 


7664 


7672 


7679 


7686 


7694 


7701 


1 1 2 


3 


4 


4 


5 


6 7 


59 


7709 


7716 


7723 


7731 


7738 


7745 


7752 


7760 


7767 


7774 


1 1 2 


3 


4 


4 


5 


6 7 


60 


7782 


7789 


7796 


7803 


7810 


7818 


7825 


7832 


7839 


7846 


1 1 2 


3 


4 


4 


5 


6 6 


61 


7853 


7860 


7868 


7875 


7882 


7889 


7896 


7903 


7910 


7917 


1 1 2 


3 


4 


4 


5 


6 6 


62 


7924 


7931 


7938 


7945 


7952 


7959 


7966 


7973 


7980 


7987 


1 1 2 


3 


3 


4 


5 


6 6 


63 


7993 


8000 


8007 


8014 


8021 


8028 


8035 


8041 


8048 


8055 


1 1 2 


3 


3 


4 


5 


5 6 


64 


8062 


8069 


8075 


8082 


8089 


8096 


8102 


8109 


8116 


8122 


1 1 2 


3 


3 


4 


5 


5 6 


65 


8129 


8136 


8142 


8149 


8156 


8162 


8169 


8176 


8182 


8189 


1 1 2 


3 


3 


4 


5 


5 6 


66 


8195 


8202 


8209 


8215 


8222 


8228 


8235 


8241 


8248 


8254 


1 1 2 


3 


3 


4 


5 


5 6 


67 


8261 


8267 


8274 


8280 


8287 


8293 


8299 


.8306 


8312 


8319 


1 1 2 


3 


3 


4 


5 


5 6 


68 


8325 


8331 


8338 


8344 


8351 


8357 


8363 


837C 


8376 


8382 


1 1 2 


3 


3 


4 


4 


5 6 


69 


8388 


8395 


8401 


8407 


8414 


8420 


8426 


8432 


8439 


8445 


1 1 2 


2 


3 


4 


4 


5 6 


70 


8451 


8457 


8463 


8470 


8476 


8482 


8488 


8494 


8500 


8506 


1 1 2 


2 


3 


4 


4 


5 6 


71 


8513 


8519 


8525 


8531 


8537 


8543 


8549 


8555 


8561 


8567 


1 1 2 


2 


3 


4 


4 


5 5 


72 


8573 


8579 


8585 


8591 


8597 


8603 


8609 


8615 


8621 


8627 


1 1 2 


2 


3 


4 


4 


5 5 


73 


8633 


8639 


8645J8651 


8657 


8663 


8669 


8675 


8681 


8686 


1 1 2 


2 


3 


4 


4 


5 5 


74 


8692 


8698 


8704 8710 


8716 


8722 


8727 


873S 


8739 


8745 


1 1 2 


2 


3 


4 


4 


5 5 


75 


8751 


8756 


87628768 


8774 


8779 


8785 


8791 


8797 


8802 


1 1 2 


2 


3 


3 


4 


5 5 


76 


8808 


8814 


8820 8825 


8831 


8837 8842 


8848 


8854 


8859 


1 1 2 


2 


3 


3 


4 


5 5 


77 


8865 


887118876 8882 


8887 


8893 8899 


8904 


8910 


8915 


1 1 2 


2 


3 


3 


4 


4 5 


78 


8921 


8927 


8932 


8938 


8943 


8949 


8954 


8960 


8965 


8971 


1 1 2 


2 


3 


3 


4 


4 5 


79 


8976 


8982 


8987 


8993 


8998 


9004 


9009 


9015 


9020 


9025 


1 1 2 


2 


3 


3 


4 


4 5 


80 


9031 


9036 


9042 


9047 


9053 


9058 


9063 


9069 


9074 


9079 


1 1 2 


2 


3 


3 


4 


4 5 


81 


9085 


9090 


9096 


9101 


9106 


9112 


9117 


9122 


9128 


9133 


1 1 2 


2 


3 


3 


4 


4 5 


82 


9138 


9143 


9149 


9154 


9159 


9165 


9170 


9175 


9180 


9186 


1 1 2 


2 


3 


3 


4 


4 5 


83 


9191 


9196 


9201 


9206 


9212 


9217 


9222 


9227 


9232 


9238 


1 1 2 


2 


3 


3 


4 


4 5 


84 
85 


9243 


9248 


9253 


9258 


9263 


9269 


9274 


9279 


9284 


9289 


1 1 2 


2 


3 


3 


4 


4 5 


9294 


9299 


9304 


9309 


9315 


9320 


9325 


9330 


9335 


9340 


1 1 2 


2 


3 


3 


4 


4 5 


86 


9345 


9350 


9355 


9360 


9365 


9370 


9375 


9380 


9385 


9390 


1 1 2 


2 


3 


3 


4 


4 5 


87 


9395 


9400 


9405 


9410 


9415 


9420 


9425 


9430 


9435 


9440 


1 1 


2 


2 


3 


3 


4 4 


88 


9445 


9450 


9455 


9460 


9465 


9469 


9474 


9479 


9484 


9489 


1 1 


2 


2 


3 


3 


4 4 


89 


9494 


9499 


9504 


9509 


9513 


9518 


9523 


9528 


9533 


9538 


1 1 


2 


2 


3 


3 


4 4 


90 


9542 


9547 


9552 


9557 


9562 


9566 


9571 


9576 


9581 


9586 


1 1 


2 


2 


3 


3 


4 4 


91 


9590 


9595 


9600 


9605 


9609 


9614 


9619 


9624 


9628 


9633 


1 1 


2 


2 


3 


3 


4 4 


92 


9638 


9643 


9647 


9652 


9657 


9661 


9666 


9671 


9675 


9680 


1 1 


2 


2 


3 


3 


4 4 


93 


9685 


9689 


9694 


9699 


9703 


9708 


9713* 


9717 


9722 


9727 


1 1 


2 


2 


3 


3 


4 4 


94 


9731 


9736 


9741 


9745 


9750 


9754 


9759 


9763 


9768 


9773 


1 1 


2 


2 


3 


3 


4 4 


95 


9777 


9782 


9786 


9791 


9795 


9800 


9805 


9809 


9814 


9818 


1 1 


2 


2 


3 


3 


4 4 


96 


9823 


9827 


9832 


9836 


9841 


9845 


9850 


9854 


9859 


9863 


1 1 


2 


2 


3 


3 


4 4 


97 


9868 


9872 


9877 


9881 


9886 


9890 


9894 


9899 


9903 


9908 


1 1 


2 


2 


3 


3 


4 4 


98 


9912 


9917 


9921 


9926 


9930 


9934 


9939 


9943 


9948 


9952 


1 1 


2 


2 


3 


3 


4 4 


99 9956 
1 


9961 


9965 


9969 


9974 


9978 


9983 


9987 


9991 


9996 


1 1 


2 


2 


3 


3 


3 4 



209 



2IO APPENDIX 

Table IX 
REFERENCES ON ENGINEERING THERMODYNAMICS 

Engineering Thermodynamics. Charles E. Lucke. McGraw-Hill Book Co., 

New York City, N. Y. 
Heat Engineering. Arthur M. Greene, Jr. McGraw-Hill Book Co., New York 

City, N. Y. 
Applied Thermodynamics. Wm. D. Ennis. D. Van Nostrand Co., New York 

City, N. Y. 
Principles of Thermodynamics. G. A. Goodenough. H. Holt & Co., New York 

City, N. Y. 
Thermodynamics of the Steam Engine and Other Heat Engines. Cecil H. Peabody. 

John Wiley & Sons, Inc., New York City, N. Y. 
Elements of Heat Power Engineering. Hirshfeld and Barnard. John Wiley & 

Sons, Inc., New York City, N. Y. 
Thermodynamics. Dr. Max Planck, Alex. Ogg, Trans. Longmans, Green & Co., 

New York City, N. Y. 
Thermodynamics. De Volson Wood. John Wiley & Sons, Inc., New York City, 

N. Y. 
Technical Thermodynamics. Gustav Zeuner, J. F. Klein, Trans. D. Van Nos- 
trand Co., New York City, N. Y. 
Thermodynamics of Heat Engines. Sidney A. Reeve. MacMillan Company, 

New York City, N. Y. 
An Introduction to General Thermodynamics. Henry A. Perkins. John Wiley & 

Sons, Inc., New York City, N. Y. 
Thermodynamics of Technical Gas Reactions. Dr. F. Haber. Longmans, Green 

& Co., New York City, N. Y. 
The Thermodynamic Principles of Engine Design. L. M. Hobbs. C. Griffin & 

Co., London. 
An Outline of the Theory of Thermodynamics. Edgar Buckingham. MacMillan 

Company, New York City, N. Y. 
The Temperature-Entropy Diagram. Charles W. Berry. John Wiley & Sons, Inc., 

New York City, N. Y. 
The Steam Engine and Other Steam Motors. Robert C. H. Heck. D. Van 

Nostrand Co., New York City, N. Y. 
The Steam Engine and Turbine. Robert C. H. Heck. D. Van Nostrand Co., 

New York City, N. Y. 
Steam Engine, Theory and Practice. Wm. Ripper. Longmans, Green & Co., 

New York City, N. Y. 
Internal Combustion Engines. R. C. Carpenter, and Diederichs. D. Van Nos- 
trand Co., New York City, N. Y. 
Gas, Petrol and Oil Engines, Vol. I. Dugald Clerk. John Wiley & Sons, Inc., 

New York City, N. Y. 
Gas, Petrol and Oil Engines, Vol II. Clerk and Burls. John Wiley & Sons, Inc., 

New York City, N. Y. 
The Design and Construction of Internal Combustion Engines. Hugo Guldner. 

D. Van Nostrand Co., New York City, N. Y. 



APPENDIX 211 

Gas Engine Design. C. E. Lucke. D. Van Nostrand Company, New York City, 

N. Y. 
Modern Gas Engine and the Gas Producer. A. M. Levin. John Wiley & Sons, 

Inc., New York City, N. Y. 
The Gas Turbine. H. Holzwarth. C. G. Griffin Company, London. 
Physical Significance of Entropy. J. F. Klein. D. Van Nostrand Co., New York 

City, N. Y. 
Practical Treatise on the "Otto" Cycle Gas Engine. Wm. Norris. Longmans, 

Green & Co., New York City, N. Y. 
Steam Tables and Diagrams. Marks & Davis. Longmans, Green & Co., New 

York City, N. Y. 
Properties of Steam and Ammonia. G. A. Goodenough, John Wiley & Sons, Inc., 

New York City, N. Y. 
Tables of the Properties of Steam and Other Vapors and Temperature Entropy 

Table. Cecil H. Peabody. John Wiley ^ Sons, Inc., New York City, N. Y. 
Thermodynamic Properties of Ammonia. Robert B. Brownlee, Frederick G. 

Keyes. John Wiley & Sons, Inc., New York City, N. Y. 
Reflections on the Motive Power of Heat and on Machines fitted to Develop Power. 

N. L. S. Carnot. Edited by R. H. Thurston, John Wiley & Sons, Inc., New 

York City, N. Y. 
Energy. Sidney A. Reeve. McGraw-Hill Book Co., New York City, N. Y. 
A New Analysis of the Cylinder Performance of Reciprocating Engines. J. P. 

Clayton, University of Illinois. Univ. of 111. Bulletin, Vol. IX, No. 26. 

Urbana, 111. 191 2. 
Thermal Properties of Steam. G. A. Goodenough. Bulletin of the University of 

Illinois, Urbana, 111. Vol. XII, No. 1. 1914. 



INDEX 



Absolute temperature, 3, 15. 

pressure, 6. 

zero, 3. 
Absorption system of refrigeration, 1 
Adiabatic expansion, 27, 32, 36, 43, 
no, 128. 

compression, 32, 36, 43. 

lines of steam, 109. 
Air compressor, 178. 

engines, 49. 
Ammonia machine, 191. 
Apu, 67. 
Available energy of steam, 128-134. 

Barrel calorimeter, 84. 
Barrus' calorimeter, 81, 82. 
Boyle's law, 15, 29. 
Brake horse power, 8. 
Brayton cycle, 57. 
British thermal unit, 3. 

Calorie, 3. 

Calorimeter, steam, 77-85. 

Carnot, 40. 

Carnot cycle, 40, 117. 

cycle, efficiency of, 44. 

cycle, entropy changes, 94. 

cycle, reversed, 45. 
Centigrade, 2, 3. 
Charles' law, 16. 
Clausius, 119. 

cycle, 119. 
Clayton's analysis, 147. 
Coefficient of flow, 165. 

of performance, 194. 
Combined diagrams, 140. 
Combination law, 16. 
Compound engine, 140. 
Compressed air, 178-185. 
Compression, adiabatic, 32, 36, 43. 



Compression, isothermal, 27, 42. 
of gases, 26. 
of vapors, 104. 

93. Compressor, air, 178. 

94, Condensing calorimeters, 84. 
Conservation of energy, 9. 
Conversion of temperatures, 2, 3. 
Cycle, Brayton, 57. 

Carnot, 40. 

Clausius, 119. 

definition of, 40. 

hot-air engine, 49. 

internal-combustion engine, 53. 

Otto, 56. 

Rankine, 119, 140. 

reversible, 45. 

steam engine, 125. 

Stirling, 50. 

Density, 65, 154, 198. 

Diagram, temperature-entropy, 95, 134. 

heat-entropy, in. 

indicator, 26, 47, 140. 

Mollier, 99, 113, (Appendix). 

total heat-entropy, 1 1 1 , 113. 
Discharge, maximum, 135. 
Dry saturated steam, 71. 
Drying of steam by throttling, 75. 

Efficiency, air engines, 52, 53. 

Carnot cycle, 44. 

Ericsson engine, 53. 

Rankine cycle, 119, 140. 

refrigerating machine, 94. 

Stirling engine, 52. 

thermal, n, 140. 

volumetric, 181, 183. 
Energy, available, 128-134. 

internal, 21, 35, 69. 

intrinsic, 69. 
213 



214 



INDEX 



Engine, Ericsson, 52. 

compound, 140. 

hot-air, 49. 

Lenoir, 53. 

Stirling, 50. 
Entropy, 91. 

diagram, 94-102. 

of the evaporation, 97. 

of the liquid, 96. 
Equivalent evaporation, 85. 
Ericsson hot-air engine, 52. 
Evaporation, equivalent, 85. 

external work of, 67. 

factor of, 85. 

internal energy of, 69. 

latent heat of, 67. 
Expansion of gases, 26. 

of vapors, 104. 
Expansions, adiabatic, 27, 32, 36, 43. 

isothermal, 27, 28, 42, 94. 

poly tropic, 112. 
External work, 67. 

work in steam formation, 67. 

Factor of evaporation, 85. 
First law of thermodynamics, 9. 
Fliegner's formulas, 155. 
Flow of air, 155. 

in nozzles, 150. 

in orifices, 150 

of steam, 167, 171. 
Foot-pound, 6. 
Foot, square, 6. 

Gas constant (it), 18, 22. 
Gas, perfect, 14, 18. 
Gram-calorie, 3. 
Grashof, 168. 

h (heat of liquid), 66. 

H (total heat of steam), 68. 

Heat engine efficiencies, 40-60. 

of liquid, 66. 

units, 3. 
Heat-entropy diagram, 111. 
Hirn's analysis, 144. 



Hot-air engine, 49. 

Hyperbolic logarithms, 31, (Appendix). 

Horse power, 8. 

Indicator diagram, 26, 47, 140. 

diagram, combined, 140. 
Injectors, 165. 

efficiency of, 166. 
Internal energy, 21, 35, 69. 

of evaporation, 69. 
Intrinsic energy, 69. 
Isothermal expansion, 27, 28, 42, 94. 

compression, 27, 42. 

lines of steam, 108. 

Joule's law, 21. v 

Kelvin, 21. 

L (latent heat of steam), 67. 
Latent heat of evaporation, 67. 
Laws of perfect gases, 18. 

of thermodynamics, 9. 
Logarithms, natural, 31, (Appendix). 

"common" or ordinary, 31. 
Losses, turbine, 163. 

Mean specific heat, 74, 75. 
Mechanical equivalent of heat, 8. 
Moisture in steam, 72, 77. 
Mollier diagram, 99, 113 (Appendix). 

Naperian logarithms, 31, (Appendix). 
Napier's formula, 168. 
Natural logarithms, 31, (Appendix). 
Non-expansive cycle, 127. 
Non-reversible cycle, 45. 
Nozzle, flow through, 150. 
Nozzle losses, 173. 
Nozzles, impulse, 162. 
reaction, 164. 

Orifice, flow through, 150. 
Otto cycle, 56. 
Per cent wet, 72. 
Perfect gas, 14, 18. 



INDEX 



2I 5 



Porous plug experiment, 21. 
Power plant diagrams, 100. 
Pressure-temperature relation, 16, 36. 

units, 6. 
Properties of gases, 197. 

of steam, 63. 

of vapors, 62. 

g, 66. 

Quality of steam, 72, no, 113. 

R (thermodynamic or "gas" constant), 

18, 22, 48. 
r, 67. 

Rankine cycle, 119, 140. 
Ratio of expansion (r), 30. 
of specific heats, 23, 154. 
Receiver method, 158. 
References, 198. 
Refrigerating machines, 40, 178. 

media, 87. 
Refrigeration, applications of, 178. 

unit of, 187. 
Regenerator, 51. 
Reversibility, 45. 
Reversible cycle, 45. 

Saturated steam, 62, 65. 
Saturation curve, 142. 
Scales, the rmo metric, 2, 3. 
Second law of thermodynamics, 9. 
Separating calorimeter, 82. 
Small calorie, 3. 
Specific heat, 4, 23. 

heat at constant pressure, 5, 197. 

heat at constant volume, 5, 197. 

heat, instantaneous values of, 76. 

heat, mean value of, 75. 

heat, ratio of, 23. 

heat of gases, 197. 

heat of superheated steam, 75, 76. 

heat of water, 4. 

volume of gases, 6, 198. 

volume of saturated steam, 65. 

volume of superheated steam, 75. 
Steam, dry, 71. 



Steam engines, efficiencies of, 119. 

engines, power plant, 100. 

entropy for, 96. 

saturated, 71. 

superheated, 72, 73. 

tables, (Appendix). 

total heat of, 68. 

turbine, 163. 

wet, 71. 
Stirling engine, 50. 
Superheated steam, 72. 
Superheating calorimeter, 77, 132. 
Symbols, ix, x. 

Tables, steam, (Appendix). 

vapor, 63. 
Temperature, absolute, 3. 

-entropy diagrams, 95-102, 134. 

-entropy diagram for power plant, 
100. 
Thermal efficiency, n, 140. 

unit, British, 3. 
Thermodynamics, definition of, 1. 
Thermo metric scales, 2, 3. 
Throttling, 75. 

Throttling calorimeter, 77, 80, 81. 
Total heat-entropy diagram, in, 113. 

heat of saturated steam, 68. 

heat of superheated steam, 73. 

internal energy of steam, 69. 
Turbine, steam, 163. 

Under-expansion, 173. 
Unit of heat, 3. 

Vaporization (see evaporation). 
Vapors, 62, 104. 

properties of, 62. 
Velocity, 155. 

Volume, specific, 6, 65, 75, 198. 
Volumetric efficiency, 181, 183. 

Water, specific heat of, 4. 
Watt, 8. 

Weight, units of, 3 (footnote). 
Wet steam, 71, no, 131. 



2l6 



INDEX 



Wetness of steam, 72. 
Wire-drawing, 75. 
Work, external, 6, 67. 

of adiabatic expansion, 67. 

of Carnot cycle, 44. 

of Clausius cycle, 120. 



Work, of compression, 179. 
of Rankine cycle, 120. 

x (quality of steam), 72, no. 

Zero, absolute, 3. 




TOTAL HEAT-ENTROPY DIAGRAM. 

The ordinates are Total Heats. 

The abscissae are Entropies. 

Vertical lines are lines of constant entropy. 

Horizontal lines are lines of constant total heat. 

Reproduced by permission from 
Marks and Davis' Steam Tables. 



S 



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